Work Energy and Power Test

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Work Energy and Power Test - 51

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Question 1
1 / -0

Velocity-time graph of a particle of mass $$2\,\,kg$$ moving in a straight line as shown in figure. Work done by all the forces on the particle is :

A
$$400\,J$$

B
$$- 400\,J$$

C
$$- 200\,J$$

D
$$200\,J$$

Solution

From the graph we have
at t =0
Velocity of particle $$(V_1) = 20 \, ms^{-1}$$
$$\Rightarrow KE_1 = \frac {1}{2}mv_1^2 = \frac {1}{2}\times 2 \times 20 \times 20$$
$$= 400\, J$$
Now,
at t = 2 sec.
Velocity of particle $$(V_2)=0$$
$$\Rightarrow KE_2=0$$
From, work - energy theorem
Work done by all forces (W) is equal to the change in kinetic enery $$(\Delta KE)$$ of the body.
i.e. $$W = \Delta KE$$
$$\Rightarrow W = KE_2 - KE_1$$
$$=0-400$$
$$\Rightarrow W= - 400 \, J$$
Question 2
1 / -0

A force $$F=Kx^{2}$$ acts on a particle at angle of $$60^{o}$$ with $$x-$$ axis. The work done in displacing the particle from $$x_{1}$$ to $$x_{2}$$ will be:

A
$$\dfrac{kx^{2}}{2}$$

B
$$\dfrac{k}{6}(x_{3}^{2}-x_{1}^{3})$$

C
$$\dfrac{k}{2}(x_{2}^{3}-x_{1}^{3})$$

D
$$\dfrac{k}{3}(x_{2}^{3}-x_{1}^{3})$$

Solution
Question 3
1 / -0

A body of mass $$6\,kg$$ is under a force which causes displacement in it given by $$S = \dfrac{{{t^2}}}{4}$$ metres where $$t$$ is time. The work done by the force in $$2$$ seconds is:-

A
$$12\,J$$

B
$$9\,J$$

C
$$6\,J$$

D
$$3\,J$$

Solution

we have
displacement ($$s$$) given as
$$s=\cfrac{{t}^{2}}{4}$$
so, velocity $$v=\cfrac{ds}{dt}$$
i.e $$v=\cfrac{2t}{v}$$
$$v=\cfrac{t}{2}{ms}^{-1}$$
Now at $$t=0$$. velocity $${v}_{1}=0$$
and hence kinetic energy $${k}_{1}=0$$
at $$t=2sec$$
velocity $${v}_{2}=\cfrac{2}{2}=1{ms}^{-1}$$
So, kinetic energy, $${k}_{2}=\cfrac{1}{2}m{v}_{2}^{2}=\cfrac{1}{2}\times 6\times {(1)}^{2}$$
$$=3J$$
By work energy theorem
Work done by the force $$=$$ change in kinetic energy
i.e., $$w=\Delta K$$
$$w={K}_{2}-{K}_{1}=3-0$$
Work done $$($$w$$)=3$$ joule
Question 4
1 / -0

A light particle moving horizontally with a speed of $$12\ m/s$$ strikes a very heavy block moving in the same direction at $$10\ m/s$$. The collision is one-dimensional and elastic. If the speed of the block does not change after the collision, the particle will

A
Move at $$2\ m/s$$ in its original direction

B
Move at $$8\ m/s$$ in its original direction

C
Move at $$8\ m/s$$ opposite to its original direction

D
Move at $$12\ m/s$$ opposite to its original direction

Solution

Conservation of momentum,

$$m_1v_1+m_2v_2=m_1v_1'+m_2v_2'$$

$$\Rightarrow m_1(v_1-v_1')=m_2(v_2-v_2')$$.......(1)

Conservation of kinetic energy requires,

$$\dfrac{1}{2}(m_1v_1^2+m_2v_2^2)=\dfrac{1}{2}(m_1{{v_1'}^2}+m_2{{v_2'}^2})$$

$$\Rightarrow m_1(v_1^2-{{v_1'}^2})=m_2(v_2^2-{{v_2'}^2})$$

$$\Rightarrow m_1(v_1-v'_1)(v_1+v'_1)=m_2(v_2-v'_2)(v_2+v'_2)$$.........(2)

dividing (2) and (1)

$$v_1+v'_1=v_2+v'_2$$

$$12+v'_1=10+10$$

$$v'_1=8m/s$$
Question 5
1 / -0

A body of mass 'M' collides against a wall with a velocity v and retraces its path with the same speed. the change in momentum is ............. (take initial direction of velocity as positive)

A
Zero

B
2Mv

C
Mv

D
-2 Mv

Solution

Taking +x direction to be positive , and assuming ball was travelling in +x direction initially.
$$P_i=Mv$$
After collision ball will move in -x direction
$$P_f=-Mv$$
Change in momentum:
$$\Delta P= P_i-P_f $$
$$\Delta P= Mv+Mv=2Mv$$
Question 6
1 / -0

A spring with spring constant $$k$$ is extended from $$x=0$$ to $$x={x}_{1}$$. The work done will be

A
$${kx}_{1}^{2}$$

B
$$\dfrac{1}{2}{kx}_{1}^{2}$$

C
$$2{kx}_{1}^{2}$$

D
$$2{kx}_{1}$$

Solution

we know that
work done = change in Enegy
$$W=\Delta E$$
and energy of spring is defined as $$\dfrac{1}{2}kx^2$$ where $$x$$ is stretch in spring.
$$\therefore W=\dfrac{1}{2}kx_1^2$$
Question 7
1 / -0

A body of mass $$5 kg$$ under a force which causes displacement in it given $$S=\dfrac{t^2}{4}$$ meter where $$'t'$$ is time. The work done by the force in $$4 seconds$$ is:

A
$$12J$$

B
$$20 J$$

C
$$2.55 J$$

D
$$10 J$$

Solution

After $$2s$$
$$\Rightarrow s=\dfrac{t^2}{4}$$

$$\Rightarrow s=\dfrac{2^2}{4}=1m$$

$$\Rightarrow v=\dfrac{ds}{dt}$$

$$\Rightarrow v=\dfrac{2t}{4}$$
$$v=\dfrac{t}{2}$$

$$\Rightarrow a=\dfrac{dv}{dt}$$
$$=\dfrac{1}{2}m/s^2$$
$$\Rightarrow F=ma$$
$$=5\times \dfrac{1}{2}$$
$$\Rightarrow F=2.5N$$
$$\Rightarrow w=F\times s$$
$$=2.5\times 1$$
$$=2.55$$
Work done by the force in 4 seconds is $$2.55.$$
Question 8
1 / -0

Five equal force of $$10$$N each applied at one point and all are lying in one plane. If the angles between them are equal, the resultant force will be

A
zero

B
$$10$$N

C
$$20$$N

D
$$10\sqrt2$$

Solution

It is clear that all the vectors are inclined at equal plane w.r.t each other. Also since all the forces lie in the same plane the resultant will be a zero vector would cancel each other. Even in real life if we pull an object with the same force from different positions with equal angles between each force the object would not move.
If mathematically calculated then,
Angle between the forces $$Q=\dfrac { { 360 } }{ 5 } ={ 72 }^{ 0 }$$
Resultant can be calculated $${ F }_{ R }=\dfrac { F\sin\dfrac { N\theta }{ 2 } }{ \sin\dfrac { \theta }{ 2 } } $$
On Substituting we get $${ F }_{ R }=0$$.
Question 9
1 / -0

A body is moving along y-axis and force acting on it is given by F = sin Ky, where K is a constant . The work done by the force from y = 0 to y = 1 is

A
$$\dfrac { 1 }{ K(1-sinK) } $$

B
$$\dfrac { { 2sin }^{ 2 }\dfrac { K }{ 2 } }{ K } $$

C
$$\dfrac { cosK-1 }{ K } $$

D
cos K - 1
Question 10
1 / -0

When the mass of body is halved and velocity is doubled, then the kinetic energy of the body

A
remains same

B
is doubled

C
is 4 times

D
is $$\frac{1}{4}$$ th

Solution

$$K.E = \dfrac{1}{2}mv^2$$

$$New\ K.E = \dfrac{1}{2} (\dfrac{m}{2})\ (2v)^2$$ = $$2 \times \dfrac{1}{2}mv^2$$ = $$2 \times K.E$$

Hence, kinetic energy of body is doubled.