[5 Mark Question Answer]

Question 15 Marks

(i) (a) Define work.
(b) What are the conditions for doing work?
(c) State the mathematical expression for work,
(ii) A force of 20 N is required to pull up a body of mass 1.5 kg through a distance of 6 m along an inclined plane as shown in diagram alongside. Calculate
(a) Force of gravity acting on the body. (Take $g =10 ms^{-2}$ )
(b) Work done by the force in pulling body along the inclined plane.
(c) Work done against the force due to gravity.
(d) Account for the difference in answers (b) and (c)

(iii) A body has a KE ' P '. If the mass of body increases 25 times, calculate its velocity, the kinetic energy remaining same.

Answer

(i) (a) Work is said to be done only when the force applied on a body causes a displacement of the body in the direction of force or any resolved component of the force.
(b) Conditions for doing work :
(i) There must be a force acting on the body.
(ii) The force acting on the body must cause a displacement.
(iii) The displacement must be in the direction of force or its resolved component.
(c) Mathematically, work is the product of force and the displacement in the direction of force.
$W=\overrightarrow{F} \overrightarrow{S}=FS \cos \theta$
(ii) (a) Force of gravity acting on the body $=m g=1.5 kg \times 10 ms^{-2}=15 N$.
(b) Work done by the force in pulling body along inclined plane
$=\text { Force } \times \text { displacement }=20 N \times 6 m=120 J .$
(c) Work done against force of gravity $=$ Force of gravity $\times$ vertical height $=m g \times BC$
$=m g \times AC \sin 30^{\circ}=15 N \times 6 m \times \frac{1}{2}=45 J .$
(d) The difference in work done $(120 J-45 J)=75 J$ is the work done against the force of friction between the body and the inclined plane.
(iii) Let initial velocity =$V$
$\therefore KE=\frac{1}{2} m V^2$...(i)
When mass increases 25 times, let velocity be $V_1$,
$\therefore KE=\frac{1}{2} \times 25 m V_1^2$....(ii)
As KE remains the same,
$\therefore \frac{1}{2} \times 25 m V_1^2=\frac{1}{2} m V^2 \Rightarrow 25 V_1^2=V^2 \text { or } V_1=\frac{V}{5}$

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Question 25 Marks

(i) What is potential energy. Discuss about two types of potential energy.
(ii) (a) State law of conservation of energy.
(b) Write two examples of interconversion of energy.
(iii) A body A of mass 20 kg is moving with a velocity of $1 ms^{-1}$. Another body B of mass 1 kg is moving with a velocity of $20 ms^{-1}$. Find the ratio of kinetic energy of $A$ and $B$.

Answer

(i) Potential Energy :
If a body does work by virtue of its position on configuration, it is said to have potential energy.
Generally potential energy is of two types, namely :
(a) Gravitational potential energy (b) Blastic potential Energy
(a) Gravitational potential Energy : The potential energy possessed by a body due to its position is called gravitational potential energy.
(b) Elastic potential Energy : Elasticity is the property by virtue of which a body regains its original shape after removal of the external force. The potential energy possessed by a body due to change in its configuration is called the elastic potential energy.
(ii) (a) Energy can neither be created nor destroyed. It may be transformed from one form to another form, but the total energy of the system remains constant.
(b) (i) When knife is rubbed against grinding stone, the mechanical energy changes into heat, light and sound energy.
(ii) The electrical energy in an electromagnet changes in
(iii) K.E. of $A =\frac{1}{2} m_1 V_1^2=\frac{1}{2} \times 20 kg \times\left(1 ms^{-1}\right)^2=10 J$
K.E. of $B =\frac{1}{2} m_2 V_2^2=\frac{1}{2} \times 1 kg \times\left(20 ms^{-1}\right)^2=200 J$
Ratio of K.E. of $A$ and $B$ is (K.E. $)_A:(K . E)_B=10: 200=1: 20$.

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Question 35 Marks

$(i) (a)$ Define work and state the mathematical expression for work.
$(b)$ State $\text{CGS}$ and $SI$ unit of work.
$(ii)$ An electric pump is $60 \%$ efficient and is rated $2 HP$ . Calculate the maximum amount of water it can lift through a height of $5 m$ in $40 s .\ [$Take $g =10 ms^{-2}$ and $1 HP =750 W$ ]
$(iii)$ In a hydroelectric power station, $1000 \ kg$ water is allowed to drop through a height of $100 m$ in $1s$ . If the conversion of potential energy to electric energy is $60 \%$, calculate the power output in $kW ($take $g =10 m / s ^2).$

Answer

$(i) (a)$ Work is said to be done only when the force applied on a body causes a displacement of the body in the direction of force or any resolved component of the force.
Mathematically, work is the product of force and the displacement in the direction of force.
$W=\overrightarrow{F} \cdot \overrightarrow{S}=FS \cos \theta$
$(b) \ \text{CGS}$ unit of work $- \text{erg}$
$SI$ unit of work $- $joule
$(ii)$ Energy supplied to the motor $= P \times t$
$ 2 HP \times 40 s=2 \times 750 W \times 40 s=1500 W \times 40 s=6 \times 10^4 J$
Useful work done by the motor $=6 \times 10^4 J \times \frac{60}{100}=36000 J$
Useful work done in lifting water $=m \ \text{mgh}$
$=m \times 10 ms^{-2} \times 5 m$
$\Rightarrow m \times 50 m^2 s^{-2}=36000 J$
$\therefore m=\frac{36000 J}{50 m^2 s^{-2}}=720 \ kg$
$(iii) \ P.E$. of water $= \text{mgh} =1000 \ kg \times 10 ms^{-2} \times 100 m=10^6 J$
Useful $P.E. =10^6 J \times \frac{60}{100}=6 \times 10^5 J$
$\therefore$ Electrical power output $=\frac{ W }{t}=\frac{6 \times 10^5 J}{1 s}$
$=6 \times 10^5 W$
$=6 \times 10^2 kW=600 kW$.

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Question 45 Marks

(i) State fhe law of conservation of energy.
(ii) Prove mathematically the law of conservation of energy.
(iii) Explain how a freely swinging pendulum obeys the law of conservation of energy.

Answer

(i) Law of conservation of energy : Energy can neither be created, nor can it be destroyed. It can be transformed from one form to another form. But the total energy of a system remains constant.
(ii) Mathematical proof : At point $A$ : Consider a body of mass $m$

at rest at a height ' $h$ ' from the ground.
P.E. at the height $h=m g h ;$ K.B. at the height $h=0$
The total energy possessed by the body at A = mgh
At point B : At point ' B ' the height from the ground $=h-x$
It $V$ be the velocity acquired by the body at B.
Applying, $V^2-u^2=2 g h$
$\Rightarrow V^2-0^2=2 g x$
$\therefore V^2=2 g x$
Then,
$\begin{aligned}\text { K.B. at } B & =\frac{1}{2} m V^2 \\
& =\frac{1}{2} m(2 g x)=m g x \\
& =m g(h-x)\end{aligned}$
P.E. at B
Total energy at $B =m g x+m g(h-x)=m g h$
At point C : At the point ' C ' on the surface, let the velocity be ' $V$ ' Applying, $V^2-u^2=2 g h$
$\Rightarrow V^2-0^2=2 g h \quad(\because$ Body is rest at height $h)$
$\therefore V^2=2 g h$
Now, K.E., at $C =\frac{1}{2} m V^2=\frac{1}{2} m(2 g h)=m g h$
P.E. at $C=m g(0)=0$
Total energy at $C =m g h+0=m g h$
So, total energy of the freely falling body at the three points $=m g h$.
Thus, the total energy of a system is constant.
(iii) In case of swinging pendulum :
-At the extreme position C, P.E. is maximum and K.E. $=0$.
-At the mean position A , the pendulum has maximum velocity and its height is lowest. Thus, the potential energy is zero and K.E. is maximum.
-At position B, the pendulum gains maximum height ' $h$ ' and the velocity is zero. Thus, P.E. is maximum and K.E. is zero.
Thus, for the earth and pendulum system, the P.E. contimuously transforms into K.E. and vice-versa and the total energy remains constant.

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Physics [5 Mark Question Answer]

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