208 questions · self-marked practice — reveal the answer and mark yourself.
Explanation:
Kinetic energy is possessed by a body by virtue of its state of motion. So, a body at rest cannot possess kinetic energy. A body at rest will possess potential energy. Thermal and magnetic energies are irrespective of state of rest or of motion of a body.
Explanation:
Moving vehicle on the road possesses kinetic energy.
A running athlete possesses kinetic energy.
Stone on the road possesses kinetic energy.
A stretched rubber band possesses potential energy.
Explanation:
When a ball is projected upwards it's height increases.
As height increases, v velocity decreases (Kinetic Energy Decreases) so Potential energy increases.
Potential Energy = mgh
Explanation:
We know that, P.E = mgh
So, It is directly proportional to height and mass.
Since both the weights are at the same height, so the weight with a larger mass will have more potential energy.
Since 10kg object has a larger mass than a 5 medical history
So, the potential energy of a 10kg mass will be greater.
Explanation:
When the force on a particle is always perpendicular to its velocity, the work done by the force on the particle is zero, as the angle between the force and velocity is 90°. So, kinetic energy of the particle will remain constant. The force acting perpendicular to the velocity of the particle provides centripetal acceleration that causes the particle to move in a circular path.
Solution:
As power, P = force × veclocity
$\text{P}=\big[\text{MLT}^{-2}\big]\big[\text{LT}^{-1}\big]=\big[\text{ML}^2\text{T}^{-3}\big]$
As, $\text{P}=\big[\text{ML}^2\text{T}^{-3}\big]$
= Constant
$\therefore\text{ L}^2\text{T}^3=\text{Constant}$
Or, $\frac{\text{L}^2}{\text{T}^3}=\text{Constant}$
$\therefore\text{ L}^2\propto\text{T}^3$
Or, $\text{L}\propto \text{T}^{\frac{3}{2}}$
Hence, right choice is (iii) $\text{t}^{\frac{3}{2}}$
Explanation:
Taking + x direction to be positive, and assuming ball was travelling in + x direction initially.
Pi = Mv
After collision ball will move in - x direction
Pf = − Mv
Change in momentum:
ΔP = Pi − Pf
ΔP = Mv + Mv = 2Mv
Explanation:
When you compress a spring, it possess potential energy. The force of compression is proportional to the compression, according to Hooke's Law. Releasing the spring turns the potential energy into kinetic energy. The spring can be then used to propel some object.
Explanation:
Suppose that one end of an extensible string is attached to a mass m, while the other end is fixed. The mass moves with a velocity v in a vertical circle of radius R. At some instant, the string makes an angle $\theta$ with the vertical as shown in the figure.
For a complete circle, the minimum velocity at L must be $\nu_\text{L}=\sqrt{5\text{gl}}.$
Applying the law of conservation of energy, we have:
Total energy at M = total energy at L
i.e., $\frac{1}{2}\text{m}\nu_{\text{M}^2}+\text{mgl}=\frac{1}{2}\text{m}\nu_{\text{L}^2}$
$\Rightarrow\frac{1}{2}\text{m}\nu_{\text{M}^2}=\frac{1}{2}\text{m}\nu_{\text{L}^2}-\text{mgl}$
Using $\nu_\text{L}\geq\sqrt{5\text{gl}},$ we have:
$\frac{1}{2}\text{m}\nu_{\text{M}^2}\geq\frac{1}{2}\text{m}(5\text{gl})-\text{mgl}$
$\therefore\ \nu_\text{M}=\sqrt{3\text{gl}}$
Conservative forces are those forces in which work done (i) in a closed path is zero and (ii) is independent of path.
Conservative forces: Gravitational and Electrostatic force.
Non-conservative forces: Frictional force and air resistance.
Explanation:
The energy exerted or work done by our muscles on the rubber band is consumed in changing its shape. It gets stored in the stretched rubber band as its potential energy. It is this stored energy that is used by the rubber band to move to its original state, shape, and size. When the rubber band is released, this stored potential energy gets converted into kinetic energy.
If a pebble is placed in contact with the stretched rubber band, this kinetic energy is transferred to the pebble. This kinetic energy of the pebble is enough to do some destructive work, like breaking a glass window, injuring someone, etc.
Explanation:
As the distance keeps on decreasing and there will be a deceleration in the blocks.
In further results, collision increases frequently many times and slowly come in contact.
Explanation:
When a stone is placed at the roof it is at a certain height that is given by In the = m × g × h
As we have a certain value for h there would be some value for potential energy.
As the stone is at rest at the roof it will not have any kinetic energy as its velocity is zero, $\text{K}=\frac{1}{2}(\text{m}\times\text{in}^2)=0$
Explanation:
Here, a ball of mass m moving with velocity in collides elastically with wall hence, momentum is.
pi = min
The ball rebounds from wall hence, final momentum is.
pf = −min
Change in momentum is.
Δp = pi − pf
Δp = mv − (−min) = 2min
Explanation:
The potential energy of a two particle system depends only on the separation between the particles.
Explanation:
By definition, $\text{p}\frac{\text{dW}}{\text{dt}}$
$\because$ Work done = Kinetic energy
$\Rightarrow\text{p}=\frac{\text{dW}}{\text{dt}}$
$=\frac{\text{d(KE)}}{\text{dt}}=\text{constant}$
For an elastic spring, the work done is equal to the negative of the change in its potential energy.
When the spring was initially compressed or stretched by a distance x, its potential energy is given by,
$(\text{P.E.})_\text{i}=\frac{1}{2}\text{kx}^2$
When it finally comes to its natural length, its potential energy is given by,
$(\text{P.E.})_\text{f}=0$
$\therefore$ Work done $=-[(\text{P.E.})_\text{f}-(\text{P.E.})_\text{i}]=-\Big[0-\frac{1}{2}\text{kx}^2\Big]$
$=\frac{1}{2}\text{kx}^2$
Explanation:
No, the earth does not perform any work on the moon.
Work done(W) is defined as the scalar product of force(F) and displacement(s).
So, W = F × s = FsCosI where is the angle between force and displacement vector.
The radial force exerted on the moon by earth i.e attractive force due to gravity acts in direction perpendicular to which the moon suffers the displacement during rotation.
So, I = 90 hence CosI = 0
So, W = |F|| s| × 0 = 0
Explanation:
We know that Potential energy is the energy possessed by a body by the virtue of its position.
$\therefore$ An overhead tank having some water possesses Potential energy, as it is at a height.
Explanation:
The potential energy possessed by the object is the energy present in it by virtue of its position or configuration. So, the hanging coconut has potential energy due to its location (height).
Explanation:
Explanation:
When external forces acting on a system have zero resultant, the centre of mass may move with a constant velocity i.e. it must not accelerate.
The path taken by the suitcase.
The time taken by you in doing so.
Your weight.
Work done by us on the suitcase is equal to the change in potential energy of the suitcase.
i.e., W = mgh
Here, mg is the weight of the suitcase and h is height of the table.
Hence, work done by the conservative (gravitational) force does not depend on the path.
Explanation:
The potential energy possessed by the object is the energy present in it by virtue of its position or configuration. Water stored in a dam, when allowed to flow, has kinetic energy which was earlier stored as potential energy.
Explanation:
Kinetic energy is due to the motion of the object and potential energy is due to relative elevation.
So as the plane takes off height increases and also its speed increase, so both kinetic and potential energy increases.
Kinetic energy$=\frac{1}{2}\text{min}^2$
Potential energy = mgH
Explanation:
Consider that the stone is projected with initial speed v.
As the stone is falls under the gravitational force, which is a conservative force, the total energy of the stone remains the same at every point during its motion.
From the conservation of energy, we have:
Initial energy of the stone = final energy of the stone
i.e., (K.E.)i + (P.E.)i = (K.E.)f + (P.E.)f
$=\frac{1}{2}\text{mv}_\text{r}^2+\text{mgh}=\frac{1}{2}\text{m}(\text{v}_\text{max})^2$
$\Rightarrow\text{v}_\text{max}=\sqrt{\text{v}^2+2\text{gh}}$
From the above expression, we can say that the maximum speed with which the stone hits the ground depends on the speed of projection and greater than it.
Explanation:
Conservation of Momentum principle, Initial momentum is Mi = 5m1
Final momentum is Mf = 4(m1 + m2)
By the above stated principle Mi = Mf = 5m1 = 4(m1 + m2)
⟹ m1 = 4m2
$\therefore$ m1 : m2= 4 : 1
Explanation:
When work is done by an external forces on a system, the total energy of the system will change.
Explanation:
mass of girl M = 50kg
h = 1. 2m
A girl is jumping vertically upward, when it will reach at maximum Hight its velocity will become zero ie inf = 0
K. And $=\frac{1}{2}\text{min}^2=\frac{1}{2}\text{m}\times0=0$
Explanation:
Potential energy is stored in a wound spring. Potential energy is a type of mechanical energy. Hence, energy present in a wound spring is mechanical energy.
Explanation:
$\text{power}=\frac{\text{Work}}{\text{Time}}=\frac{\text{mgh}}{\text{t}}$
Here, m = 100kg, h = 10m
and t = 1 min = 60s
$\therefore\text{p}=\frac{100\times9.8\times10}{60}=163.3\text{W}$
Explanation:
The sphere is rolling on a smooth surface i.e., no energy is lost to friction and the kinetic energy of sphere A is transferred to the sphere B without any loss as the collision is head-on and elastic.
Thus, the kinetic energy of sphere A and sphere B are equal their ratio will be 1 : 1
Explanation:
When an arrow is drawn back by a bow, the work done by us in stretching the bowstring gets stored at potential energy in the bow.
This potential energy of bow is transformed into kinetic energy when the bowstring is released and this gives kinetic energy to the arrow.
Explanation:
Constant after some depth. This constant velocity is called terminal velocity, hence KE become constant beyond this depth, which is best represented by (b).
Explanation:
The energy which a body possesses by virtue of being in motion is called as kinetic energy.
If we want to stop a moving body then we must do some work on the body to stop it which means that a moving body possesses energy because of motion.
If an object of mass m moving with the velocity v, The kinetic energy is defined as,
$\text{K.E}=\frac{1}{2}\text{mv}^2$
Explanation:
Mass of the boy, m = 40kg
Number of steps = 50
Height of each step = 10cm
Force on the boy due to gravity, F = mg = 40 × 9. 8 N = 392N
While climbing up the steps, the boy does work against gravity.
Displacement in the vertical direction, s = (50 × 10)cm = 500cm = 5m
Displacement is in the direction of force applied by the boy against gravity.
So, work done, In = F × s = 392 × 5J = 1960J
If the collision is elastic, which of the following is a possible result after collision?
Explanation:
Key concept: In a collision if the motion of colliding particles before and after the collision is along the same line, the collision is said to be head on or one dimensional.
When two bodies of equal masses collides elastically, their velocities are interchanged. Kinetic energy and linear momentum remains conserved Total kinetic energy of the system before collision
$=\frac{1}{2}\text{mv}^2+0=\frac{1}{2}\text{mv}^2$
In (a), kinetic energy of the system after collision,
$\text{K}_1=\frac{1}{2}(2\text{m})\Big(\frac{\text{v}}{2}\Big)^2=\frac{1}{4}\text{mv}^2$
Hence this option is incorrect.
In (b), kinetic energy of the system after collision,
$\text{K}_2=\frac{1}{2}(\text{m})(\text{v})^2=\frac{1}{2}\text{mv}^2$
Hence this option will be correct.
In (c), kinetic energy of the system after collision,
$\text{K}_3=\frac{1}{2}(3\text{m})\Big(\frac{\text{v}}{3}\Big)^2=\frac{1}{6}\text{mv}^2$
Hence this option is incorrect.
In (d), kinetic energy of the system after collision,
$\text{K}_4=\frac{1}{2}\text{mv}^2+\frac{1}{2}\text{m}\Big(\frac{\text{v}}{2}\Big)^2+\frac{1}{2}\text{m}\Big(\frac{\text{v}}{3}\Big)^2=\frac{49}{72}\text{mv}^2$
We see that kinetic energy is conserves only in (b).
Explanation:
The kinetic energy of a body is the energy by virtue of its motion. This is the definition of kinetic energy. If it has motion then it will have velocity, hence kinetic energy exists.
$\text{K.E}=\frac{1}{2}\text{mv}^2$
Explanation:
We know that force $\text{F}=\frac{\Delta\text{p}}{\Delta\text{t}}$ and $\frac{\Delta\text{p}}{\Delta\text{t}}=\frac{\text{m}[(-\text{v})-\text{u}]}{\text{t}}=\frac{-2\text{mv}}{\text{t}}$
And the magnitude of force will be, $\text{F}=\frac{2\text{mv}}{\text{t}}$
According to the problem, $\text{m}=150\text{g}=\frac{150}{1000}\text{kg}=\frac{3}{20}\text{kg}$
$\Delta\text{t}=\text{time of contact}=0.001\text{s}$
$\text{u}=126\text{km/ h}=\frac{126\times1000}{60\times60}\text{m/ s}=35\text{}\text{m s}$
$\text{v}=-126\text{km/ h}=-35\text{m/ s}$
Change in momentum of the ball
$\Delta\text{P}=\text{m(v}-\text{u})$
$=\frac{3}{20}(-70)=-\frac{21}{2}$
We know that force $\text{F}=\frac{\Delta\text{P}}{\Delta\text{t}}$
$=\frac{\frac{-21}{2}}{0.001}\text{N}=-1.05\times10^4\text{N}$
Here, -ve sign shows that force will be opposite to the direction of movement of the ball before hitting. So the force that the batsman had to apply to hold the bat firmly at its place would be F = 1.05 × 104N K
Explanation:
Kinetic energy is due to motion of the object and potential energy is due to relative position of object wrt earth.
In option A. Relative position is same so no potential energy if we assume reference as level of road( assuming horizontal road), but have kinetic energy as the man is in motion.
In option C he is lying on bed so neither relative change in position nor body in motion so have none of the energies kinetic or potential.
In option D this is same case as that of option A. (assuming level park).
But in option B, man is climbing a hill so relative position wrt ground level changes so it gain some potential energy and also he is moving so also have kinetic energy.
Explanation:
W = F.S → 4(4 − 1)J = 12J
Explanation:
If u and v are the velocities before collision and u' and v' are the velocities after collision, then we have
$\text{u}'=\frac{(\text{m}-\text{m})\text{u}}{\text{m}+\text{m}}+\frac{2\text{m}}{\text{m}+\text{m}}=0+\text{v}=\text{v}$ and $\text{v}'=\frac{2\text{m}\text{u}}{(\text{m}+\text{m})}+\frac{(\text{m}-\text{m})\text{v}}{(\text{m}+\text{m})}=\text{u}+0=\text{u}$
So the velocities and speeds are interchanged. Hence (a) and (b) are true.
Since the velocities are interchanged and masses are equal hence the momenta are also interchanged. Hence (c) is true.
If u > v then after the collision the speeds of bodies are interchanged. Now the faster body slows down and the slower body speeds up. Hence (d) is true.
Explanation:
Water is rich in hydrogen (proton). On collision, velocities of neutron and proton are interchanged. Fast neutrons come to rest and protons move with velocity of neutrons.
Explanation:
Kinetic energy is defined as the energy possessed by a body by virtue of its motion. and it is equal to K. And, $\text{uh}=\frac{1}{2}\text{min}^2$
Where m and in are mass and Velocity of moving body.
Explanation:
Two forces act on the bob. (i) is the tension in the string and the (ii) mg sin I, which will be tangential to the path.
The resultant of both the forces will be along vector C
Explanation:
Tension in the cord is maximum (for a given average speed of rotation) when the mass, m, is at the bottom points B, as the gravitational force is in the downward direction and tension of the cord is directly opposing it.
Explanation:
As the collision is inelastic, body losses some energy, so that KE of ball does not remain the same. However, total energy and total momentum of ball and earth remain the same.
Explanation:
The work done by the spring on both the masses is equal to the negative of the increase in the elastic potential energy of the spring.
The elastic potential energy of the spring is given by $\text{E}_\text{p}=\frac{1}{2}\text{kx}^2.$
Work done by the spring on both the masses $=-\frac{1}{2}\text{kx}^2$
$\therefore$ Work done by the spring on each mass $=\frac{1}{2}\Big(-\frac{1}{2}\text{kx}^2\Big)$
$=-\frac{1}{4}\text{kx}^2$
Explanation:
According to the principle of conservation of linear momentum,
$\text{m}_1\text{v}_1-\text{m}_2\text{v}_2=0$
$\therefore\frac{\text{m}_1}{\text{m}_2}=\frac{\text{v}_2}{\text{v}_1}$
$\frac{\text{E}_1}{\text{E}_2}=\frac{\frac{1}{2}\text{m}_1\text{v}_1^2}{\frac{1}{2}\text{m}_2\text{v}_2^2}$
$=\frac{\text{m}_1}{\text{m}_1}\Big(\frac{\text{v}_1}{\text{v}_2}\Big)^2=\frac{\text{m}_1}{\text{m}_2}\Big(\frac{\text{m}_2}{\text{m}_1}\Big)^2$
$=\frac{\text{E}_1}{\text{E}_2}=\frac{\text{m}_2}{\text{m}_1}$
Explanation:
Potential energy is the energy possessed due to the relative position of one body with respect to other.
Kinetic energy is the energy possessed due to the motion of the body.
Water stored in the reservoir is at rest, so no kinetic energy but it is at some height with respect to some level below the water surface, so it contains potential energy.
Explanation:
Key concept: To calculate the change in kinetic energy of the system during motion we have to apply work energy, theorem. According to this theorem, Net work done = Final kinetic energy - Initial kinetic energy of the object The above statement shows the connection between work and kinetic energy as:
“The work done by the net force acting on an object is equal to the change in the kinetic energy of that object”.
Net work done (IF) on a particle equals change in kinetic energy of the particle.
$\sum\text{W}=\text{K}_2-\text{K}_1$
According to the problem as the electron and proton are moving under the influence of mutual forces, the magnetic forces will be perpendicular to their motion, hence, it acts as a centripetal force for the particle. In this way the particle performs the uniform circular morion, this implies speed will remain constant. So, there is no change in kinetic energy of the particle. Hence no work is done by these forces.
$\vec{\text{F}_\text{m}}=\text{q}(\vec{\text{v}}\times\vec{\text{B}})\cdot\text{F}_\text{m}$
(magnetic force) will be perpendicular to both B and v, where B is the external magnetic field and v is the velocity of particle. That is why one ignores the magnetic force of one particle on another.
Explanation:
In the elastic collision between a heavy object and a very light object at rest, the velocity of particles after collision is.
for heavy particle, in1 = in the1
for light particle, in2 = 2in the1 −in the2
since, in the2 = 0 hence,
in2 = 2in the1
Therefore, the stone will fly away with a velocity equal to
in2 = 2in the1 = 2(10) = 20ms−1
Explanation:
F = 16N
A = 8 × 8 × 10−4m2
$\text{P}=\frac{\text{F}}{\text{A}}=\frac{16}{64 \times10^{-4}}=\frac{1000}{4}=2500\text{Pa}$
Explanation:
P.E. decreases when an air bubble rises in water. Because work is done by upthrust.
Explanation:
Since, $\text{P}=\frac{\text{W}}{\text{t}}$
So, if W is constant, than $\text{P}\propto\frac{1}{\text{t}}$
i.e. $\frac{\text{P}_1}{\text{P}_2}=\frac{\text{t}_2}{\text{t}_1}=\frac{20}{10}$
$\Rightarrow\frac{\text{P}_1}{\text{P}_2}=\frac{2}{1}$
P1 : P2
= 2 : 1
Explanation:
As the masses of two balls are equal, their velocities are exchanged. As first ball comes to rest, speed of other ball = v.
Explanation:
As the child stand up the effective length of pendulum decreases due to the reason that center of gravity rises up.
According to $\text{T}=2\text{s}\sqrt{\Big(\frac{1}{\text{g}}}\Big)$
Explanation:
A raised hammer possesses P.E Energy possessed by a body by virtue of its position is called as potential energy.
Similarly is the case when a hammer is raised. A raised hammer possesses potential energy by virtue of its height above ground level.
Explanation:
The work done by a person in lifting an object is stored as its potential energy mgh.
Hence, work done depends on the weight of the object mg.
Explanation:
As shown in diagram AB and AC are two smooth planes inclined to the angle $\theta_1$ and $\theta_2$ respectively. As friction is absent here, hence, mechanical energy will be conserved. As both the tracks having common height h, From conservation of mechanical energy,
$\frac{1}{2}\text{mv}^2=\text{mgh}$ (for both tracks I and II)
$\text{v}=\sqrt{2\text{gh}}$
Hence, speed is same for both stones. For stone I, a1 = acceleration along inclines plane $=\text{g}\sin\theta_1$
Similarly, for stone II, $\text{a}_2=\text{g}\sin\theta_2$ as $\theta_2>\theta_1$ hence, $\text{a}_2>\text{a}_1$
And both length for track II is also less, hence stone II reaches earkier than stone I.
Explanation:
Acceleration of the block will be the same to both the observers. The respective kinetic energies of the observers are different, because the block appears to be moving with different velocities to both the observers. Work done by the friction and the total work done on the block are also different to the observers.
Explanation:
Chemical energy is energy stored in the bonds of atoms and molecules. Batteries, biomass, petroleum, natural gas, and coal are examples of chemical energy. Chemical energy is converted to thermal energy when people burn wood in a fireplace or burn gasoline in a car's engine.
Explanation:
$\triangle\text{p=2mvcos}60^\circ=\text{min}=2×3×10×21=30\frac{\text{kgm}}{\text{s}}$
Explanation:
The net force on the block is not zero, therefore the block will not be in any given equilibrium.
Explanation:
In uniform circular motion speed is constant while velocity being a vector quantity is constantly changing as its direction keeps changing. Centripetal force acts inwards towards the center to counterbalance the centrifugal force acting outwards from the center.
Explanation:
As work is done by the spring, therefore, there are only two possibilities: the spring was initially compressed by a distance x and has come to its natural length or the spring was initially stretched by distance x and finally comes to its natural length.
Explanation:
The change of kinetic energy will be equet to the change in potential energy.
$\triangle\text{KE}=\triangle\text{PE}=\text{mg}(\text{h}_1-\text{h}_2)$
Since radious r,
h1 - h2 = r
Explanation:
Tension in the string is equal to F, as tension on both sides of a frictionless and massless pulley is the same.
i.e., T - Mg = Ma
⇒ T = Mg + Ma
So, the tension in the string cannot be equal to Mg.
The change in kinetic energy of the block is equal to the work done by gravity.
Hence, the work done by gravity is 20J in 1s, while the the work done by the tension force is zero.
Explanation:
Given, mass m1 = m2 = m
and velocity, v = v1
For elastic collision, $\text{v}_2=\Big(\frac{\text{m}_2-\text{m}_1}{\text{m}_1+\text{m}_2}\Big)\text{v}_2+\frac{2\text{m}_1\text{v}_1}{\text{m}_1+\text{m}_2}$
After putting given values, we will get
$\text{v}_2+\frac{2\text{m}_1\text{v}_1}{2\text{m}_1}$
$\Rightarrow\text{v}_1=\text{v}_2$
Explanation:
For simple pendulum
$\frac{\text{min}}{\text{r}}=\text{T}-\text{mg}\cos\varnothing$
But when F = I, i.e., the bob is at extreme position. its velocity is zero, hence the equation becomes: $\text{T−mg}\cosθ=0$
$\Rightarrow\text{T−mg}\cosθ=0$
Explanation:
Positron because their charges are same. As the mass of positron is much lesser than proton, (1/1840 times) it moves away through much larger distance compared to proton. Change in their momentum will be same. So, velocity of lighter particle will be greater than that of a heavier particle. So, positron is moved through a larger distance.
As work done = force × distance. As forces are same in case of proton and positron but distance moved by positron is larger, hence, work done will be more.
Explanation:
Key concept: Energy Graph for a Spring: If the mass attached with spring performs simple harmonic motion about its mean position then its potential energy at any position (x) can be given by,
Now kinetic energy at any position $\text{K}=\text{E}-\text{U}=\frac{1}{2}\text{Ka}^2-\frac{1}{2}\text{Kx}^2$
$\text{K}=\frac{1}{2}(\text{a}^2-\text{x}^2)\ ....(\text{i})$
From the above formula, we can cheak that
$\text{U}_\text{max}=\frac{1}{2}\text{Ka}^2$ $\big[\text{At extreme x}=\pm\text{a}\big]$ and Umin = 0 [At mean x = 0]
$\text{K}_\text{max}=\frac{1}{2}\text{Ka}^2$ [At mean x = 0] and Kmin = 0 $\big[\text{At extreme x}=\pm\text{a}\big]$
$\text{E}=\frac{1}{2}\text{ka}^2=\text{constant(at all positions)}$
Because velocity of mass = 0 [at extreme position]
$\therefore\ \text{K}=\frac{1}{2}\text{mv}^2=0$
It means kinetic energy changes parabolically w.r.t. position but total energy remain always constant irrespective to the position of mass.
Total energy is E = PE + KE, which remains constant.
According to the question, when paraticle is at x = xm, i.e., at extreme position x = xm ⇒ KE = 0
Hence, total energy will be
$\text{E}=\text{PE}+0=\text{PE}$
$\Rightarrow\text{V}\text{(x}_\text{m})=\frac{1}{2}\text{kx}^2_\text{m}$
Hence option (b) is correct.
V = E, K = 0
Explanation:
When the bob is swinging in vertical circle the centripetal force and the pseudo force varies.
Centripetal force causes the tension in the string.
Let centripetal force be C, pseudo force be P and weight of bob be In.
At position 1, C = P − In
At position 2, C = In
At position 3, C = P + In
As C at position 3 is greatest, the tension in the string is greatest at this position.
Explanation:
When a body is placed at a certain height, the body possesses potential energy.
P. And, uh. = mgH
m; the mass of the object
g; acceleration due to gravity
H; height gained by object
Explanation:
Work done,
In = Fd
Displacement d is given by,
$\text{d}=\frac{1}{2}\text{at}^2$
F = ma
10 = 10a, a = 1m/ s2
$\text{d}=\frac{1}{2}(1)(2)^2=2\text{m}$
In = 10 × 2 = 20D
Explanation:
Work is area under the curve.
So In = In1 + In2 + In3 + In4
In1 = arandaundandr A1BCM2
In2 = arandaundandr M2DEF3
In3 = arandaundandr F3GHI4
In4 = arandaundandr I4JKL5
In1 = 10 × 1 = 10ergs
In2 = 20 × 1 = 20ergs
In3 = −20 × 1 = −20ergs
In4 = 10 × 1 = 10ergs
In = In1 + In2 + In3 + In4 = 10 + 20 − 20 + 10 = 20ergs
Explanation:
As the friction is present in fhis problem, so mechanical energy is not conserved. So energy will be lost due to dissipation by friction^Here, work is done by the frictional force on the cycle and is equal to 200 × 10 = -2000J
As the road does not move at all, therefore, work done by the cycle on the road is zero. Important point. We should be aware that here the energy of bicyclist is lost during the motion, but it is lost due to friction in the form of heat.
Explanation:
We will use Energy mass equivalence
E = mc2
E = 10−6kg × (3 × 108)2m/ s2
E = 9 × 1010J
Explanation:
When m1 = m2 the statements are true.
Explanation:
When a pendulum oscillates in air, its total mechanical energy decreases continuously in overcoming resistance due to air. Therefore, total mechanical energy of the pendulum decreases exponentially with time. The variation of E v/st is correctly represented by curve (c) in which the relation between energy and time is shown.
Explanation:
At height h from ground raindrop have maximum potential energy. And kinetic velocity will be zero (at the instant when it dropped its velocity will be zero), then as the rain drop falls its PE starts decreasing and kinetic energy start increasing.
The total mechanical energy will remain conserved if we neglect the air resistance. If there is some air resistance, there is some force called upthrust (in fluids) which opposes its motion. It depends upon velocity of object as the velocity increases, upthrust also increases. Hence during fall of raindrop first its velocity increases and then become constant after some time. This constant velocity is called terminal velocity, hence KE also become constant. PE decreases continuously as the drop is falling continuously. The variation in PE and KE is best represented by (b).
Explanation:
work done = mgh
m = mass of drop = 0.1g = 0.00001kg
g = 9.8m/ s
h = 100m
work done = 0.00001 × 9.8 × 10
= 0.098 J
Explanation:
v = u + at
15 = 6 + a(2)
a = 4.5m/ s2
s = ut + 0.5at2 = 6(2) + 0.5(4.5)(4) = 21m
W = mas = 2(4.5)(21) = 189J
Explanation:
mcarucar + mtruckutruck = mcarvcar + mtruckvtruck
400 × 72 + 4000 × 9 = −18 × 400 + 4000 × v
v = 18kmph
Explanation:
We know that total energy is kinetic energy plus potential energy.
Now, kinetic energy is,
And $=\frac{1}{2}\text{Min}^2$
Which depends on velocity. In watch there is no displacement in the knob, hence velocity is zero. So there is no kinetic energy. Only potential energy is there.
Explanation:
Potential energy is energy stored in an object. This energy has the potential to do work. Gravity gives potential energy to an object. This potential energy is a result of gravity pulling downwards. The gravitational constant, g, is the acceleration of an object due to gravity. This acceleration is about 10 meters per second on earth. The formula for potential energy due to gravity is PE = mgh. As the object gets closer to the ground, its potential energy decreases while its kinetic energy increases.
Explanation:
Potential energy of a body is defined as energy of a body due to its position in gravitational field.
In general, Potential energy = M × g × h where: h is the height above ground.
If the person will be lying on ground then it will have minimum height above the ground therefore potential energy of the person will also be minimum.
Explanation:
At any time force acting on particle vary and hence acceleration (net) will have different direction at different times.
Tension also changes and its minimum at top point. Magnitude of acceleration also varies.
Considering earth and particle as a system and no external force on system is acting, total mechanical energy will be conserved.
Explanation:
$1\text{KE}_2=\frac{1}{2}\text{KE}_1$
$\frac{1}{2}\text{mv}^2_2=\frac{1}{2}\cdot\frac{1}{2}\text{mv}^2_1$
$\text{v}^2_2=\frac{1}{2}\text{v}^2_1=\Big(\frac{\text{v}_1}{\sqrt{2}}\Big)^2$
$=\Big(\frac{\text{v}_1\sqrt{2}}{2}\Big)^2=(0.707\text{v}_1)^2$
Explanation:
Hydro power plant uses the potential energy stored in water. When water flows down the dam, potential energy is converted into kinetic energy which is used to rotate the turbine which produce electricity.
Explanation:
Key concept: When a variable force acts on a particle while it moves from point A to B, say along the path shown in the figure, work done by the force on the particle is given by
On the particle is given by
$\text{W}=\int\limits^{\text{B}}_\text{A}\vec{\text{F}}\cdot\vec{\text{ds}}\ ...(\text{i})$
Hence, $\int\limits^{\text{B}}_\text{A}\vec{\text{F}}\cdot\vec{\text{ds}}$ is to be integrated along the path the particle follows.
The vector integral is equivalent to
$\text{W}=\int\limits^{\text{x}_2}_{\text{x}_1}\text{f}_\text{x}\text{dx}+\int\limits^{\text{y}_2}_{\text{y}_1}\text{f}_\text{y}\text{dy}+\int\limits^{\text{z}_2}_{\text{z}_1}\text{f}_\text{z}\text{dz}$
According to the problem, velocity = ax3/2, mass = 0.5kg, a = 5m-1/2s-1
We have to find work done (W) by net force.
We know that,
Acceleration, $\text{a}_0=\frac{\text{dv}}{\text{dt}}=\text{v}\frac{\text{dv}}{\text{dx}}$
$=\text{ax}^{\frac{3}{2}}\frac{\text{d}}{\text{dx}}\Big(\text{ax}^{\frac{3}{2}}\Big)$
$=\text{ax}^{\frac{3}{2}}\times\text{a}\times\frac{3}{2}\times\text{x}^{\frac{1}{2}}=\frac{3}{2}\text{a}^2\text{x}^2$
Net Force $=\text{ma}_0=\text{m}\Big(\frac{3}{2}\text{a}^2\text{x}^2\Big)$
And work done due to variable force,
Work done $=\int\limits^{\text{x}=2}_{\text{x}=0}\text{F dx}=\int\limits^2_0\frac{3}{2}\text{ma}^2\text{x}^2\text{ dx}$
$=\frac{3}{2}\text{ma}^2\times\Big(\frac{\text{x}^3}{3}\Big)^2_0$
$=\frac{1}{2}\text{ma}^2\times8$
$=\frac{1}{2}\times(0.5)\times(25)\times8=50\text{J}$
Explanation:
Work done, W = force × displacement = mg × h = mgh
This is the potential energy.
If weight (mg) is known, then we need to know the height (h) of the table.
Explanation:
In the process of squatting on the ground he gets straight up and stand. Then he is tilted somewhat, the man exerts a variable force on the ground to balance his weight, hence he also has to balance frictional force besides his, weight in this case.
N = Normal reaction force = friction + mg ⇒ N > mg
Once the man gets straight up that variable force = 0
Normal reaction force = mg
Explanation:
Mass of water lifted, m = 500kg
Displacement, d = 80m
Time taken, t = 10s
Force, F = m × g
F = 500 × 10
F = 5000N
Work done, In = F × d
In = 5000 × 80
In = 4 × 105 J.
Explanation:
The energy possessed by a body due to its change in position or shape is called the potential energy. A wound watch has potential energy.
Explanation:
Potential energy = mgh
m → mass
g → accelerationduetogravity
h → heightfromground
P.E. is directly proportional to the height from the ground. Hence a student sitting at the top of a tree has more potential energy than the student who is sitting on the ground.
Explanation:
Earth's pull on the ball remains constant (W = mg).
Tension in the string changes along the vertical circular path.
Speed of the ball changes as sum of kinetic energy and potential energy remains constant during motion.
Speed changes, hence, centripetal force changes.
Explanation:
K. And, uh.$=\frac{1}{2}\text{min}^2$
So the kinetic energy of a body depends upon its mass and velocity.
Explanation:
Initial velocity = 10ms-1
Final velocity $=\frac{50}{10}\times2+10=20\text{ms}^{-1}$
$\Big(\text{Acceleration}=\frac{50}{10}=5\text{m/ s}^2\Big)$
Initial $\text{KE}=\frac{1}{2}\times10\times10\times10=5\times10^2\text{J}$
Final $\text{KE}=\frac{1}{2}\times10\times20\times20=20\times10^2\text{J}$
% increase $=\frac{(20-5)\times10^2}{5\times10^2}\times100=300\%$
Explanation:
Potential energy $\propto\text{x}^2$ When × becomes 5 times, P.E. becomes 25 times.
Explanation:
Potential energy P = mgh
Given, h1 = x ; h2 = 4x
Since, the masses are same,
then$\frac{\text{P}_1}{\text{P}_2}=\frac{\text{h}_1}{\text{h}_2}=\frac{\text{x}}{4\text{x}}=1:4$
Explanation:
According to the work-energy theorem, the total work done on a particle is equal to the change in kinetic energy of the particle.
Explanation:
$\text{K.E}=\frac{\text{p}^2}{2\text{m}}$
When K.E. becomes 4 times, $\text{p}^2$ is 4 times.
Therefore, p becomes 2 times.
Explanation:
The surface area of sphere < surface area of the cube for the given same mass and same density.
The rate of cooling ∝ area of contact with surroundings.
The plate will cool faster than the sphere.
Explanation:
As K.E. of particle is increasing continuously with time magnitude of its linear momentum must be increasing continuously $\big(\therefore\text{E}=\frac{\text{p}^2}{2\text{m}}\big).$ For this resultant force on the particle must be at an angle less than 90° all the time.
Explanation:
Tension at any point in vertical motion is given by:
$\text{T}=\frac{\text{min}^2}{1}+\text{mgcos}\theta$
where I = angular displacement from lowest point,
l = length of string
m = mass of string
It is clear that tension at the lowest point (B) is greatest than at other points (A, C, D). If we increase average velocity, tension will increase at lowest point, therefore at point B, string has maximum possibility of break.
Explanation:
Kinetic energy of a particle is directly proportional to the square of its velocity. The resultant force on the particle must be at an angle less than 90° with the velocity all the time so that the velocity or kinetic energy of the particle keeps on increasing.
The kinetic energy is also directly proportional to the square of its momentum, therefore it continuously increases with the increase in momentum of the particle.
Explanation:
As the body is falling freely under gravity, the potential energy decreases continuously and kinetic energy increases continuously as all the conservative forces are doing work. So, total mechanical energy (PE + KE) of the body will be constant.
Let us discuss this in detail:
In the given diagram an object is dropped from-a height H from ground.
At point A total mechanical energy will be EA = K.E + P.E
$\text{E}_\text{A}=\frac{1}{2}\text{mv}^2+\text{mgH}$
As velocity will be zero at A, so its kinetic energy will be zero.
$\text{E}_\text{A}=\text{mgH}$
Velocity at point B will be, $\text{v}_\text{B}=\sqrt{2\text{gh}}$
So, energy at point B will be $\text{E}_\text{B}=\text{KE}+\text{PE}$
$\text{E}_\text{B}=\frac{1}{2}\text{m}(2\text{gh})+\text{mg}(\text{H}-\text{h})$
$\text{E}_\text{B}=\text{mgh}+\text{mgH}-\text{mgh}$
$\text{E}_\text{B}=\text{mgH}$
Now, velocity at point C will be $\text{v}_\text{c}=\sqrt{2\text{gh}}$
So, energy at point will be $\text{E}_\text{c}=\text{KE}+\text{PE}$
$\text{E}_\text{c}=\frac{1}{2}\text{m}(2\text{gH})+\text{mg}(0)$
$\text{E}_\text{c}=\text{mgH}$
So, total mechanical energy will remain same (if we neglect the air friction).
Explanation:
Potential energy is the energy that is stored in an object due to its position above the ground surface and when the object is in motion then it has kinetic energy. Water in a flowing river has kinetic energy.
Explanation:
$\text{W}=\text{Fs}=\cos\theta=0$, when either s = 0 or $\theta=90^\circ$ i.e., when object is stationary but the point of application of the force moves on the object or object moves in such a way that point of application of force remains fixed; or force is at 90° to the acceleration.
Explanation:
According to the problem, Radius = 1m, mass = m = 5kg
$\text{f}=\frac{300}{60}$
Angular velocity will be
$=2\pi\text{f}=(300\times2\pi)\text{rad/ min}$
$=(300\times3.14)\text{rad/ 60s}$
$=\frac{300\times2\times3.14}{60}\text{rad/ s}=10\pi\text{rad/ s}$
And relation between linear velocity and angular velocity is $\text{v}=\omega\text{R}$
$=\Big(\frac{300\times2\pi}{60}\Big)(1\text{m})$
$=10\pi\text{m/ s}$
And kinetic energy $=\frac{1}{2}\text{mv}^2$
$=\frac{1}{2}\times5\times(10\pi^2)$
$=100\pi^2\times5\times\frac{1}{2}$
$=250\pi^2\text{J}$
Explanation:
If air resistance is negligible, total mechanical energy of the system will remain constant. And let us take ground as a reference where potential energy will be zero.
According to the problem, h = 1.5 m, v = 1m/ s, m = 10 kg, g = 10m s-2
Initial energy of the shotput $=(\text{PE})_\text{i}+(\text{KE})_\text{i}$
$=\text{mgh}+\frac{1}{2}\text{mv}^2$
$=10\times10\times1.5+\frac{1}{2}\times10\times(1)^2$
$=150+5=155.0\text{J}$
From conservation of mechanical energy,
$\text{(PE)}_\text{i}+\text{(KE)}_\text{i}=\text{(PE)}_\text{f}+\text{(KE)}_\text{f}$
So, final kinetic energy of the shotput is 155J
Explanation:
Since the sphere collided elastically and there was no friction there was no impulse on the sphere along the wall. The only contact force acted was normal and that obviously was perpendicular to surface. NO change in momentum parallel to wall.
Explanation:
Since the body is much heavy these won't be much change in velocity & e = 1
$\text{i.e.,}\frac{\text{v-2}}{\text{0-2}}=-1$
⇒ v = 4m/ s
Explanation:
The speed with which it hits the ground must depend upon the speed of projection and shall always be larger than the speed of projection, because potential energy of the body shall be converted into kinetic energy.
Explanation:
As the stone falls under the gravitational force, which is a conservative force, the total energy of the stone remains the same at every point during its motion.
From the conservation of energy, we have:
Initial energy of the stone = final energy of the stone
i.e., (K.E.)i+ (P.E.)i = (K.E.)f+ (P.E.)f
$=\frac{1}{2}\text{mv}^2+\text{mgh}=\frac{1}{2}\text{m}(\text{v}_\text{max})^2$
$\Rightarrow\text{v}_\text{max}=\sqrt{\text{v}^2+2\text{gh}}$
From the above expression, we can say that the maximum speed With which stone hits the ground does not depend on the initial direction.
Explanation:
The negative of the work done by the conservative internal forces on a system is equal to the changes in potential energy.
i.e. $\text{W}=-\triangle\text{ P.E.}$
Explanation:
For constant power
Displacement $\propto\text{t}^{\frac{3}{2}}$
Because, $\text{P}=\frac{\vec{\text{F}}\cdot\vec{\text{ds}}}{\text{dt}}=\vec{\text{F}}\cdot\vec{\text{v}}=\text{constant}$
$(\because$ P = constant according to the problem$)$
Now, will by dimensional analysis
[F][v] = constant
⇒ [MLT-2][LT-1] = constant
⇒ L2T-3 = constant $(\because$ mass is constant$)$
$\Rightarrow\text{L}\propto\text{T}^{\frac{3}{2}}$
$\Rightarrow\text{ Displacement (d)}\propto\text{t}^{\frac{3}{2}}$
Explanation:
The string becomes slack when the particle reaches its highest point. This is because at the highest point, the tension in the string is minimum. At this point, potential energy of the particle is maximum, while its kinetic energy is minimum. From the law of conservation of energy, we can see that the particle again passes through the initial position where its potential energy is minimum and its kinetic energy is maximum.
Explanation:
When the length of the spring is halved, its spring constant will become double.
Slope of the force displacement graph gives the spring constant (k) of spring.
If k becomes double then slope of the graph increases, i.e. Graph shifts towards force-axis.
Explanation:
The kinetic energy of a body is given by:
$\text{K}=\frac{1}{2}\text{min}^2$
Hence, Kinetic energy depends on the velocity as:
$\text{KE}\propto\text{in}^2$
Explanation:
Kinetic energy is the energy possessed by the body by virtue of its motion and potential energy is the energy possessed by the virtue of its position and shape. Energy is required to do work.
Hence work, kinetic energy, potential energy all can be measured using same unit, i.e joule.
So, force is the physical quantity here which is different from others as given by product of mass and acceleration. S.I unit of force is N (newton).
Explanation:
The variable forces are the non-constant forces that changes with maybe time, distance or any other variable.
Explanation:
Consider the box and the ball a system. As no external force acts on this system, the velocity of the centre of mass of the system remains constant.
Explanation:
The force mv2/ r directed outwards, called centrifugal force, is not a real force.
At A, mv12/ l = T1 + mg
Explanation:
Kinetic energy is not conserved in an inelastic collision.
Explanation:
Area under force displacement curve is the work done in that interval
Area under the given figure Area of surface Area of triangle. = +
Work done $= 10×100+\frac{1}{2}×10×100$
= 1000 + 500
= 1500J
Explanation:
According to work energy theorem, ΔK. And, uh.= In and here work is done by the gravitational force.
⇒ΔK. And, uh. = In = mg(2r) = 1 × 10 × 2(1) = 20D
Explanation:
When two observers are moving with respect to each other at a speed v along a straight line, acceleration of block, if any, will be same. Distance moved may be different. Therefore, work done/K.E. of the block may appear different.
Explanation:
$\because$ Work = Force × Displacement …(i)
Explanation:
For a body moving along a circular path, the centripetal force acts along the radius while the displacement is tangential, i.e. $\theta=90^\circ$, therefore, $\text{W}=\text{Fs}\cos90^\circ=0$.
Explanation:
As some energy is loss into heat in an inelastic collision, the final kinetic energy is less than the initial kinetic energy.
Expalnation:
As we know that linear momentum
$=\sqrt{2\text{mk}}$
$\Big(\therefore\text{K}=\frac{\text{P}^2}{2\text{m}}\Big)$
For same Kinetic energy, $\text{P}\propto\sqrt{\text{m}}$
$\frac{\text{P}_1}{\text{P}_2}=\sqrt{\frac{\text{m}_1}{\text{m}_2}}=\sqrt{\frac{1}{4}}=\frac{1}{2}=1:2$
Explanation:
Let xA and xB be the extensions produced in springs A and B, respectively.
Restoring force on spring A, F = kAxA ...(1)
Restoring force on spring B, F = kBxB ...(2)
From (1) and (2), we get:
kAxA = kBxB
It is given that kA = 2kB
$\therefore\ \text{x}_\text{B}=2\text{x}_\text{A}$
Energy stored in spring A:
$\text{E}=\frac{1}{2}\text{k}_\text{A}\text{x}_\text{A}^2\ \dots(3)$
Energy stored in spring B:
$\text{E}'=\frac{1}{2}\text{k}_\text{B}\text{x}_\text{B}^2=\frac{1}{2}\Big(\frac{\text{k}_\text{A}}{2}\Big)(2\text{x}_\text{A})^2$
$\therefore\ \text{E}'=2\times\Big(\frac{1}{2}\text{k}_\text{A}\text{x}_\text{A}^2\Big)=2\text{E}$ [From (3)]
Explanation:
A body of mass 20kg is held at a height of 2m above the ground means there is no displacement because of which there is no change in its potential energy.
Explanation:
To raise the car, the crane has to do work against the force of gravity.
Therefore, the force required to lift the car, F = mg = 500 × 9. 8N = 4900N
Displacement, S = vertical height raised = 4m.
$\therefore$ Work done, In = F. S = 4900 × 4J = 19600J = 19.6kJ
Explanation:
From v = u + at = 100 – 10 × 5 = 50m/s
This is the velocity at the time of explosion. According to principle of conservation of linear momentum,
$1\times50=\frac{400}{1000}\times(-25)+\frac{600}{1000}\times\text{v}$
$(50+10)=0.6\text{v}$
$\text{v}=\frac{60}{0.6}=100\text{m/s}$
The second fragment will go upwards with a speed of 100 m/s.
Explanation:
As the earth moves once around the sun in its elliptical orbit, when the earth is closest to the sun, speed of the earth is maximum, hence KE is maximum. When the earth is farthest from the sun speed is minimum hence KE is minimum but never zero and negative.
This variation of KE vs t is correctly represented by option (d).
Explanation:
No work is done by a force on an object if the force is always perpendicular to its velocity. Acceleration does not always provide the direction of motion, so we cannot say that no work is done by a force on an object if it is always perpendicular to the acceleration. Work done is zero when the displacement is zero.
In a circular motion, force provides the centripetal acceleration. The angle between this force and the displacement is 90°, so work done by the force on an object is zero.
Explanation:
Decrease in potential energy ΔU = mg(h2−h1)
ΔU = mgh = 20 × 9.8 × 0.5 = 98J
Solution:
From,
V = u + at
V = 0 + at = at
As power, P = F × V
$\therefore$ P = (ma) × at = ma2t
As m and a are constants, therefore, $\text{P}\propto\text{t}$
Hence, right choice is (ii) t.
Explanation:
As potential energy, P = mgh
Since, there is no vertical displacement so, h = 0.
Hence, potential energy changeP = mgh = 0
The moving bus has only change in its Kinetic energy.
Explanation:
1N and 4N will never give 2N
mind it, if the side lengths of a triangle is a, b, c
a + b > = c
b + c > = a
c + a > = b
Explanation:
Gravitational potential energy is given by
E P = mgh
where m is mass so heavy object means more mass and more mass means more potential energy.
Explanation:
Potential energy depends on the height at which the object is situated. Where potential energy is measured by the product of the mass of the object, gravity, and height. In this case, there is no change in height of the object.
Explanation:
The chain starts sliding, when applied force = force of friction
(due to hanging part) (between chain and table)
$\frac{1}{3}\text{mg}=\text{f}=\mu\text{R}=\mu\Big(\frac{2}{3}\text{mg}\Big)$
$\mu=\frac{1}{2}$
Explanation:
The kinetic energy of a body is the energy by virtue of its motion.
K. And $=\frac{1}{2}\text{min}^2$
where m is mass and in is velocity.
Explanation:
During an elastic collision, all of the above statements are valid.
Explanation:
$\text{K.E}=\frac{1}{2}\text{mv}^2$
$\text{K.E}\propto\text{m}\text{ &}\text{ K.E}\propto\text{v}^2$
So doubling mass will double the kinetic energy and doubling speed will make kinetic energy 4 times.
Explanation:
If in a collision kinetic energy after collision is not equal to kinetic energy before collision, the collision is said to inelastic. Coefficient of restitution 0 < e < 1 When we are considering the two bodies as system the total external force on the system will be zero. Hence, total linear momentum of the system remain conserved. Here kinetic energy appears in other forms, i.e. energy may be lost in the form of heat and sound etc. In some cases (KE)final < (KE)initial such as when initial KE is converted into intertial energy of the product (as heat, elastic or excitation) while in other cases (KE)final > (KE)initial such as when internal energy stored in the colliding particles is released.
Examples:
Explanation:
The work done by the potter is defined as the product of the force and the displacement.
Work done = force × displacement
= m × g × 10 ( Since force = mass × gravity)
= 50 × 10 × 10
= 5KJ
Explanation:
A fruit, hanging from the top branch of a tree, is at rest at a certain height from the earth"s surface.
Explanation:
Distance (d) travelled by the elevator in time t = vt
The block is not sliding on the wedge.
Then friction force $(\text{f})=\text{mg}\sin\theta$
Work done by the friction force on the block in time t is given by,
$\text{W}=\text{Fd}\cos(90-\theta)$
$\Rightarrow\text{W}=\text{mg}\sin\theta\times\text{d}\times\cos(90-\theta)$
$\Rightarrow\text{W}=\text{mgd}\sin^2\theta$
$\therefore\ \text{W}=\text{mg}\nu\text{t}\sin^2\theta$
$\frac{1-\text{ e}}{1+\text{ e}}$
Explanation:
Here, $\text{m}_1=\text{m}_2=\text{m},\text{u}_1=\text{u},\text{u}_2=0$.
Let $\text{v}_1,\text{v}_2$ be their velocities after collision.
According to principle of conservation of linear momentum.
$\text{mu}+0=\text{m}(\text{v}_1+\text{v}_2) \text{or}\text{ v}_1+\text{v}_2=\text{u}\dots\text{(i})$
By definition,
$\text{e}=\frac{\text{v}_2-\text{v}_1}{\text{u}-0} \text{or} \text{ v}_2-\text{v}_1=\text{eu}\dots\text{(ii)}$
Add (i) and (ii),
$\text{v}_2=\frac{\text{u(1+ e)}}{2}$
$\therefore\frac{\text{v}_1}{\text{v}_2}=\frac{1-\text{ e}}{1+\text{ e}}$
Explanation:
Work = Fs $\cos\theta$
Where, F is the force applied, s is the displacement, and I is the angle between the force applied and displacement.
Hence, work done is independent of time taken.
In the given cases, I = 0∘ as the force applied are in the same direction. Also, F = mg
So In = Fs = mgs
In both the cases, mass and displacement are the same.
In = 50 × 10 × 2 = 1000 D
Explanation:
As no energy is lost into heat in an elastic collision, the initial kinetic energy is equal to the final kinetic energy.
Explanation:
Net force, $=2\hat{\text{i}}+3\hat{\text{j}}-2\hat{\text{k}+}5\hat{\text{i}}-\hat{\text{j}}-3\hat{\text{k}}$
$=7\hat{\text{i}}+2\hat{\text{j}}-5\hat{\text{k}}$
Diolacement, $\text{d}=2\hat{\text{i}}-\hat{\text{j}}+3\hat{\text{k}-}\hat{\text{i}}-\hat{\text{j}}-\hat{\text{k}}$
$=\hat{\text{i}}-2\hat{\text{j}}+2\hat{\text{k}}$
Work done $=\text{F}.\text{d}=(7\hat{\text{i}}+2\hat{\text{j}}-5\hat{\text{k}}).(\hat{\text{i}}-2\hat{\text{j}}+2\hat{\text{k}})$
$=7-4-10=-7\text{ units}$.
Explanation:
If there is not specified we always consider the collision elastic. When two bodies of equal masse collides elastically, their velocities are interchanged in these types of collision.
Kinetic energy and linear momentum remain conserved. According to the above diagram when m1 comes in contact with the spring, m1 is retarded by the spring force and m2 is accelerated by the spring force.
Explanation:
Work done by weight-lifter is zero because there is no displacement. In a locomotive, work done is zero because force and displacement are mutually perpendicular to each other.
While a person holding a suitcase, work done is zero because there is no displacement.
Explanation:
In an elastic collision where the projectile is much more massive than the target, the velocity of the target particle after the collision will be about twice that of the projectile and the projectile velocity will be essentially unchanged.
Explanation:
The kinetic energy of a body of mass mm which is moving with velocity v is $\text{K}=\frac{1}{2}\text{min}^2$
Mass is always positive and if the velocity is either positive or negative, the kinetic energy is always positive due to the square of velocity.
Explanation:
According to the law of conservation of energy,
$\frac{1}{2}\text{Mv}^2=\frac{1}{2}\Big(\frac{1}{2}\text{mv}^2\Big)+\text{mgh}$
$\Rightarrow490+245+5\times9.8\times\text{h}$
$\text{h}=\frac{245}{49}=5\text{m}$
Explanation:
The potential energy is the stored energy of an object due to its position. some examples of potential energies are gravitational potential energy, Electrostatic potential energy and elastic energy etc.
A body placed at ground will have less gravitational potential energy than a body placed at some height. therefore, potential energy changes by changing the position of object.
Explanation:
The potential energy is the energy that an object has due to its position in a force field or that a system has due to the configuration of its parts. The potential energy of the car remains the same and will not change as the road is leveled and the height of the body remains the same, although its speed increases.
Explanation:
Energy stored in spring is potential energy, and it is defined as.
$\text{E}=\frac{1}{2}\text{Kx}^2$
where k is spring constant, and x is the extension/compression in spring.