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Physics 1 Marks Question

Work, Energy, and Power · Uttarakhand Board (UBSE) STD 11 Science (Uttarakhand - English Medium). 109 questions.

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Question 11 Mark
The sign of work done by a force on a body is important to understand. State carefully if the following quantities are positive or negative: Work done by the resistive force of air on a vibrating pendulum in bringing it to rest.
Answer
Negative: The resistive force of air acts in the direction opposite to the direction of motion of the pendulum. Hence, the work done is negative in this case.
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Question 21 Mark
Underline the correct alternative: In an inelastic collision of two bodies, the quantities which do not change after the collision are the total kinetic energy/ total linear momentum/ total energy of the system of two bodies.
Answer
Total linear momentum: The total linear momentum always remains conserved whether it is an elastic collision or an inelastic collision.
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Question 31 Mark
Answer carefully, with reasons: If the potential energy of two billiard balls depends only on the separation distance between their centres, is the collision elastic or inelastic? (Note: we are talking here of potential energy corresponding to the force during collision, not gravitational potential energy).
Answer
Elastic. Explanation: In the given case, the forces involved are conservation. This is because they depend on the separation between the centres of the billiard balls. Hence, the collision is elastic.
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Question 41 Mark
Underline the correct alternative: The rate of change of total momentum of a many-particle system is proportional to the external force/ sum of the internal forces on the system.
Answer
External force: If their is no internal force work on body then the external force is require to attend change in linear momentum.
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MCQ 51 Mark
A body is moving unidirectionally under the influence of a source of constant power. Its displacement in time $t$ is proportional to:
  • A
    $\text{t}^{\frac{1}{2}}$
  • B
    $\text{t}$
  • ✓
    $\text{t}^\frac{3}{2}$
  • D
    $\text{t}^2$
Answer
Correct option: C.
$\text{t}^\frac{3}{2}$
As power,$P =$ force $\times$ veclocity
$\text{P}=\big[\text{MLT}^{-2}\big]\big[\text{LT}^{-1}\big]=\big[\text{ML}^2\text{T}^{-3}\big]$
As, $\text{P}=\big[\text{ML}^2\text{T}^{-3}\big]$
$= \text{Constant}$
$\therefore\text{ L}^2\text{T}^3=\text{Constant}$
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 $(iii) \text{t}^{\frac{3}{2}}$
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Question 61 Mark
A body of mass 2kg initially at rest moves under the action of an applied horizontal force of 7N on a table with coefficient of kinetic friction = 0.1. Compute the: Work done by friction in 10 s.
Answer
M = 2kg Applied force F = 7N Coefficient of kinetic friction $\mu = 0.1$ Normal reaction is N = mg = 2 × 9.8 = 19.6N Hence, force or friction is $\text{f} = \mu\text{N} = 1.96\text{N}$ Total force = F - f = 7 - 1.96 = 5.04N
Acceleration of body is, $\text{a}=\Big(\frac{\text{F}-\text{f}}{\text{m}}\Big)$ $=\frac{5.04}{2}\simeq2.5\text{ms}^{-2}$ Displacement of body in time t is, $\text{x}=\frac{1}{2}\text{at}^2$ Int = 10s $\text{x}=\frac{1}{2}\times2.5\times10^2$ $=125\text{m}$ Work done by friction = -fx = -1.96 × 125 = -245J
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Question 71 Mark
Underline the correct alternative: Work done by a body against friction always results in a loss of its kinetic/potential energy.
Answer
Kinetic energy: The work done against the direction of friction decrease the velocity of a body. So, there is a loss of kinetic energy of the body.
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Question 81 Mark
State if the following statements is true or false. Give reasons for your answer: Total energy of a system is always conserved, no matter what internal and external forces on the body are present.
Answer
False. Explanation: The external forces on the body may change the total energy of the body.
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MCQ 91 Mark
A body is initially at rest. It undergoes one$-$dimensional motion with constant acceleration. The power delivered to it at time $t$ is proportional to:
  • A
    $\text{t}^{\frac{1}{2}}$
  • ✓
    $\text{t}$
  • C
    $\text{t}^\frac{3}{2}$
  • D
    $\text{t}^2$
Answer
Correct option: B.
$\text{t}$
From,
$V=u+a t$
$V=0+a t=a t$
As power, $\mathrm{P}=\mathrm{F} \times \mathrm{V}$
$\therefore P=(m a) \times \text { at }=m a^2 t$
As $m$ and a are constants, therefore, $\mathrm{P} \propto \mathrm{t}$
Hence, right choice is $(ii)\  t $.
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Question 101 Mark
Answer carefully, with reasons: In an elastic collision of two billiard balls, is the total kinetic energy conserved during the short time of collision of the balls (i.e. when they are in contact)?
Answer
No. Explanation: K.E. is not conserved during the given elastic collision, K.E. before and after collision is the same. Infact, during collision, K.E. of the balls gets converted into potential energy.
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Question 111 Mark
The sign of work done by a force on a body is important to understand. State carefully if the following quantities are positive or negative: Work done by an applied force on a body moving on a rough horizontal plane with uniform velocity.
Answer
Positive: Here the body is moving on a rough horizontal plane. Frictional force opposes the motion of the body. Therefore, in order to maintain a uniform velocity, a uniform force must be applied to the body. Since the applied force acts in the direction of motion of the body, the work done is positive.
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Question 121 Mark
Answer the following: Comets move around the sun in highly elliptical orbits. The gravitational force on the comet due to the sun is not normal to the comet’s velocity in general. Yet the work done by the gravitational force over every complete orbit of the comet is zero. Why?
Answer
The gravitational force is a conservative force, hence, work done by the gravitational force over one complete (closed) orbit of comet is zero.
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Question 131 Mark
A body of mass 2kg initially at rest moves under the action of an applied horizontal force of 7N on a table with coefficient of kinetic friction = 0.1. Compute the: Work done by the net force on the body in 10s.
Answer
M = 2kg Applied force F = 7N Coefficient of kinetic friction $\mu = 0.1$ Normal reaction is N = mg = 2 × 9.8 = 19.6N Hence, force or friction is $\text{f} = \mu\text{N} = 1.96\text{N}$
Total force = F - f = 7 - 1.96 = 5.04N Acceleration of body is, $\text{a}=\Big(\frac{\text{F}-\text{f}}{\text{m}}\Big)$ $=\frac{5.04}{2}\simeq2.5\text{ms}^{-2}$ Displacement of body in time t is, $\text{x}=\frac{1}{2}\text{at}^2$ Int = 10s $\text{x}=\frac{1}{2}\times2.5\times10^2$ $=125\text{m}$ Net work done = (875 - 245) = 630J
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Question 141 Mark
Answer carefully, with reasons: What are the answers to (a) and (b) for an inelastic collision?
Answer
No; Yes. Explanation: In an inelastic collision, there is always a loss of kinetic energy, i.e., the total kinetic energy of the billiard balls before collision will always be greater than that after collision. The total linear momentum of the system of billiards balls will remain conserved even in the case of an inelastic collision.
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Question 151 Mark
The sign of work done by a force on a body is important to understand. State carefully if the following quantities are positive or negative: Work done by friction on a body sliding down an inclined plane.
Answer
Negative: Since the direction of frictional force is opposite to the direction of motion, the work done by frictional force is negative in this case.
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Question 161 Mark
Answer carefully, with reasons: Is the total linear momentum conserved during the short time of an elastic collision of two balls?
Answer
Yes. Explanation: In an elastic collision, the total linear momentum of the system always remains conserved.
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Question 171 Mark
State if the following statements is true or false. Give reasons for your answer: In an inelastic collision, the final kinetic energy is always less than the initial kinetic energy of the system.
Answer
True. Explanation: In an inelastic collision, the final kinetic energy is always less than the initial kinetic energy of the system. This is because in such collisions, there is always a loss of energy in the form of heat, sound, etc.
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Question 181 Mark
State if the following statements is true or false. Give reasons for your answer: In an elastic collision of two bodies, the momentum and energy of each body is conserved.
Answer
False. Explanation: In an elastic collision, the total energy and momentum of both the bodies, and not of each individual body, is conserved.
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Question 191 Mark
State if the following statements is true or false. Give reasons for your answer: Work done in the motion of a body over a closed loop is zero for every force in nature.
Answer
False. Explanation: Work done in the motion of a body over a closed loop is zero for a conservation force only.
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Question 201 Mark
The sign of work done by a force on a body is important to understand. State carefully if the following quantities are positive or negative: Work done by a man in lifting a bucket out of a well by means of a rope tied to the bucket.
Answer
Positive: In the given case, force and displacement are in the same direction. Hence, the sign of work done is positive. In this case, the work is done on the bucket.
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Question 211 Mark
A bolt of mass 0.3kg falls from the ceiling of an elevator moving down with an uniform speed of $7ms^{-1}$. It hits the floor of the elevator (length of the elevator = 3m) and does not rebound. What is the heat produced by the impact? Would your answer be different if the elevator were stationary?
Answer
Mass of the bolt, m= 0.3kg Speed of the elevator = 7m/s Height, h = 3m Since the relative velocity of the bolt with respect to the lift is zero, at the time of impact, potential energy gets converted into heat energy. Heat produced = Loss of potential energy = mgh = 0.3 × 9.8 × 3 = 8.82J The heat produced will remain the same even if the lift is stationary. This is because of the fact that the relative velocity of the bolt with respect to the lift will remain zero.
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Question 221 Mark
Answer the following: An artificial satellite orbiting the earth in very thin atmosphere loses its energy gradually due to dissipation against atmospheric resistance, however small. Why then does its speed increase progressively as it comes closer and closer to the earth?
Answer
As an artificial satellite gradually loses its energy due to dissipation against atmospheric resistance, its potential energy decreases rapidly. As a result, kinetic energy of satellite slightly increases i.e. its speed increases progressively.
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Question 231 Mark
A $1kg$ block situated on a rough incline is connected to a spring of spring constant $100Nm^{-1}$ as shown in. The block is released from rest with the spring in the unstretched position. The block moves 10cm down the incline before coming to rest. Find the coefficient of friction between the block and the incline. Assume that the spring has a negligible mass and the pulley is frictionless.
Answer
Mass of the block, $m = 1kg$ Spring constant, $k = 100Nm^{-1}$ Displacement in the block, $X = 10cm = 0.1m$ The given situation can be shown as in the following figure.At equilibrium:

 Normal reaction, $\text{R}=\text{mg}\cos37^{\circ}$ Frictional force, $\text{f}=\mu\text{R}=\text{mg}\sin37^{\circ}$ Where, $\mu$ is the coefficient of friction Net force acting on the block $ =\text{mg}\sin 37^{\circ}-\text{f}$ $=\text{mg}\sin37^{\circ}-\mu\text{m}\cos37^{\circ}$ $=\text{mg}(\sin37^{\circ}-\mu\cos37^{\circ})$ At equilibrium, the work done by the block is equal to the potential energy of the spring, i.e., $=\text{mg}(\sin37^{\circ}-\mu\cos37^{\circ})\text{x}=\Big(\frac{1}{2}\Big)\text{kx}^2$ $1\times9.8(\sin37^{\circ}-\mu\cos37^{\circ})=\Big(\frac{1}{2}\Big)\times100\times(0.1)$ $0.602-\mu\times0.799=0.510$ $\therefore\mu=\frac{0.092}{0.799}=0.115$
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Question 241 Mark
The sign of work done by a force on a body is important to understand. State carefully if the following quantities are positive or negative: Work done by gravitational force in the above case.
Answer
Negative: In the given case, the direction of force (vertically downward) and displacement (vertically upward) are opposite to each other. Hence, the sign of work done is negative.
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Question 251 Mark
A body of mass 2kg initially at rest moves under the action of an applied horizontal force of 7N on a table with coefficient of kinetic friction = 0.1. Compute the: Change in kinetic energy of the body in 10s.
Answer
M = 2kg Applied force F = 7N Coefficient of kinetic friction $\mu = 0.1$ Normal reaction is N = mg = 2 × 9.8 = 19.6N Hence, force or friction is $\text{f} = \mu\text{N} = 1.96\text{N}$ Total force = F - f = 7 - 1.96 = 5.04N
Acceleration of body is, $\text{a}=\Big(\frac{\text{F}-\text{f}}{\text{m}}\Big)$ $=\frac{5.04}{2}\simeq2.5\text{ms}^{-2}$ Displacement of body in time t is, $\text{x}=\frac{1}{2}\text{at}^2$ Int = 10s $\text{x}=\frac{1}{2}\times2.5\times10^2$ $=125\text{m}$ Change in kinetic energy = Net work done = 630J
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Question 261 Mark
Answer the following: The casing of a rocket in flight burns up due to friction. At whose expense is the heat energy required for burning obtained? The rocket or the atmosphere?
Answer
Heat energy required for burning of casing of rocket comes from the rocket itself. As a result of work done against friction the kinetic energy of rocket continuously deceases and this work against friction reappears as heat energy.
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Question 271 Mark
Underline the correct alternative: When a conservative force does positive work on a body, the potential energy of the body increases/ decreases/ remains unaltered.
Answer
Decreases: When we work on the body then i may be possible it become in motion so if we work on body, then kinetic energy increase and potential energy decrease.
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Question 281 Mark
A body of mass 2kg initially at rest moves under the action of an applied horizontal force of 7N on a table with coefficient of kinetic friction = 0.1. Compute the: Work done by the applied force in 10s.
Answer
M = 2kg Applied force F = 7N Coefficient of kinetic friction $\mu = 0.1$ Normal reaction is N = mg = 2 × 9.8 = 19.6N Hence, force or friction is $\text{f} = \mu\text{N} = 1.96\text{N}$ Total force = F - f = 7 - 1.96 = 5.04N
Acceleration of body is, $\text{a}=\Big(\frac{\text{F}-\text{f}}{\text{m}}\Big)$ $=\frac{5.04}{2}\simeq2.5\text{ms}^{-2}$ Displacement of body in time t is, $\text{x}=\frac{1}{2}\text{at}^2$ Int = 10s $\text{x}=\frac{1}{2}\times2.5\times10^2$ $=125\text{m}$ Work done by F is W = Fx = 7 × 125 J = 875J
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Question 291 Mark
A man raises a mass m to a height 'h' and then shifts it horizontally by a length x'. What is the work done against the force of gravity?
Answer
Work done against force of gravity is mgh since no work is done in the horizontal displacement.
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Question 301 Mark
What is the source of kinetic energy of the bullet coming out of the bullet of a rifle?
Answer
The source of kinetic energy of bullet is the potential energy of the compressed spring in the loaded rifle.
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MCQ 311 Mark
A force $\Big(\frac{\text{mv}^2}{\text{r}^2}\Big)$ is acting on a body of mass m moving with a velocity $v$ in a circle of radius $r$. What is the work done by the force in moving the body over half the circumference of the circle?
  • A
    $\frac{\text{mv}^2}{\text{r}}\times\pi\text{r}$
  • B
    $\text{Zero}$
  • C
    $\frac{\text{mv}^2}{\text{r}^2}$
  • D
    $\frac{\pi\text{r}^2}{\text{mv}^2}$
Answer
Work done is zero because force is acting at $90^\circ$ to the direction of motion $\text{w}=\text{Fs}\cos90^\circ=0$
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Question 321 Mark
Can mechanical energy be negative?
Answer
As E = K + UExplanation:
E is negative if U is negative and is numerically greater than K.
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Question 331 Mark
Work done by weight of a body moving horizontally is ­________.
Answer
Work done by weight of a body moving horizontally is ­zero .
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Question 341 Mark
Calculate the power of an electric engine which can lift 20 tonne of coal per hour from a mine 180m deep.
Answer
$\text{Power}=\frac{\text{Work done}}{\text{Time taken}}=\frac{\text{mgh}}{\text{t}}$ $=\frac{20\times1000\times9.8\times180}{60\times60}$ $=9800\text{W}$ $=\text{W}=9.8\text{kW}$
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Question 351 Mark
State work-energy theorem.
Answer
According to work-energy theorem, the work done on a body is equal to the change in K.E in it.
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Question 361 Mark
Answer carefully, with reasons: In an elastic collision of two billiard balls, is the total kinetic energy conserved during the short time of collision of the balls (i.e. when they are in contact)?
Answer
No. Explanation: K.E. is not conserved during the given elastic collision, K.E. before and after collision is the same. Infact, during collision, K.E. of the balls gets converted into potential energy.
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Question 371 Mark
When force acting on a body is perpendicular to direction of motion then work done is ­­­­­­­­­________.
Answer
When force acting on a body is perpendicular to direction of motion then work done is ­­­­­­­­­Zero.
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Question 381 Mark
Give the characteristics of inelastic collision.
Answer
  1. Kinetic energy does not remain conserved.
  2. Linear momentaum of the system remains conserved.
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Question 401 Mark
Answer with reason: In an elastic collision of two billiard balls, is the total kinetic energy conserved during the short time of collision of the balls (i.e., when they are in contact)?
Answer
Total K.E. is not conserved because a part of K.E. is used in deforming the balls during that short intevral.
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Question 411 Mark
Write down characteristics of elastic collision.
Answer
  1. Kinetic energy of the system remains conserved.
  2. Linear momentaum of the system remains conserved.
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Question 421 Mark
Is it possible to exert a force which does work on a body without changing its kinetic energy. If, so give example.
Answer
Yes, When spring is compressed or when a body is pulled with a constant velocity on a rough horizontal surface.
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Question 431 Mark
The sign of work done by a force on a body is important to understand. State carefully if the following quantities are positive or negative: Work done by an applied force on a body moving on a rough horizontal plane with uniform velocity.
Answer
Positive: Here the body is moving on a rough horizontal plane. Frictional force opposes the motion of the body. Therefore, in order to maintain a uniform velocity, a uniform force must be applied to the body. Since the applied force acts in the direction of motion of the body, the work done is positive.
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Question 441 Mark
Answer the following: Comets move around the sun in highly elliptical orbits. The gravitational force on the comet due to the sun is not normal to the comet’s velocity in general. Yet the work done by the gravitational force over every complete orbit of the comet is zero. Why?
Answer
The gravitational force is a conservative force, hence, work done by the gravitational force over one complete (closed) orbit of comet is zero.
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Question 461 Mark
Does the work done in raising a box on to a platform depends upon how fast it is raised up? If not why?
Answer
No. Work done against gravity depends only on the initial and final points. The change in energy is same as the work done.
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Question 471 Mark
A body of mass 2kg initially at rest moves under the action of an applied horizontal force of 7N on a table with coefficient of kinetic friction = 0.1. Compute the: Work done by the net force on the body in 10s.
Answer
M = 2kg Applied force F = 7N Coefficient of kinetic friction $\mu = 0.1$ Normal reaction is N = mg = 2 × 9.8 = 19.6N Hence, force or friction is $\text{f} = \mu\text{N} = 1.96\text{N}$
Total force = F - f = 7 - 1.96 = 5.04N Acceleration of body is, $\text{a}=\Big(\frac{\text{F}-\text{f}}{\text{m}}\Big)$ $=\frac{5.04}{2}\simeq2.5\text{ms}^{-2}$ Displacement of body in time t is, $\text{x}=\frac{1}{2}\text{at}^2$ Int = 10s $\text{x}=\frac{1}{2}\times2.5\times10^2$ $=125\text{m}$ Net work done = (875 - 245) = 630J
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Question 481 Mark
Answer carefully, with reasons: What are the answers to (a) and (b) for an inelastic collision?
Answer
No; Yes. Explanation: In an inelastic collision, there is always a loss of kinetic energy, i.e., the total kinetic energy of the billiard balls before collision will always be greater than that after collision. The total linear momentum of the system of billiards balls will remain conserved even in the case of an inelastic collision.
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Question 491 Mark
Two bodies of masses $M_1$ and $M_2$ have equal kinetic energy. The ratio of their linear momenta is ______.
Answer
Two bodies of masses $M_1$ and $M_2$ have equal kinetic energy. The ratio of their linear momenta is $\sqrt{\frac{\text{M}_1}{\text{M}_2}}$.
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Question 501 Mark
Why is the work done by centripetal force zero?
Answer
Because centripetal force is perpendicular to the displacement of the body.
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