Work Energy and Power Test

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Work Energy and Power Test - 27

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Question 1
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A spring of force constant k is cut in two parts at its one-third length. When both the parts are stretched by same amount. The work done in the two parts will be:
(Note- Spring constant of a spring is inversely proportional to length of spring.)

A
equal in both

B
greater for the longer part

C
greater for the shorter part

D
data insufficient

Solution

As we know that spring constant k is inversely proportional to length of spring

$$k\alpha \dfrac { 1 }{ L } $$

work done = change in potential energy

$$work\quad done=\Delta PE=\dfrac { 1 }{ 2 } k{ x }^{ 2 }$$

so work done is directly proportional to k
it work done is inversely proportional to length

$$work\quad done\quad \alpha \quad k\quad \alpha \dfrac { 1 }{ L } $$

hence work done is more for small length of spring as compared to larger.
Question 2
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A pendulum bob is raised to a height, $$h$$, and released from rest. At what height will it attain half of its maximum speed?

A
$$\displaystyle\frac{3h}{4}$$

B
$$\displaystyle\frac{h}{2}$$

C
$$\displaystyle\frac{h}{4}$$

D
$$0.707h$$

Solution

By energy conservation, $$\dfrac{1}{2}mv^2_{max} = mgh$$
Let at height $$h'$$, it attain velocity $$v'=\dfrac{v_{max}}{2} = \sqrt{gh/2}$$
Now, $$mgh = \dfrac{1}{2}mv'^2 + mgh'$$

or $$mgh = \dfrac{mgh}{4} + mgh'$$

or $$mgh'=\dfrac{3}{4}mgh$$

$$\therefore h' = \dfrac{3h}{4}$$
Question 3
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A simple pendulum of length $$l$$ has maximum angular displacement $$\displaystyle \theta .$$ Then maximum kinetic energy of a bob of mass $$m$$ is

A
$$\displaystyle \frac{1}{2}mgl$$

B
$$\displaystyle \frac{1}{2}mgl\cos \theta $$

C
$$\displaystyle mgl\left ( 1-\cos \theta \right )$$

D
$$\displaystyle \frac{1}{2}mgl\sin \theta $$

Solution

At point B, $$v=0$$ and $$P.E$$ is maximum.
At point A, $$P.E= 0$$ and $$K.E$$ is maximum.
Height of bob at B, $$AP= l-lcos\theta= l(1-cos\theta)$$
Thus $$P.E= mgl(1-cos\theta)$$
Applying conservation of energy at A and B, $$K.E_m +0 = 0 + P.E$$
$$\implies K.E_m=mgl (1-cos\theta) $$
Question 4
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A particle of mass $$m$$ moves on a straight line with its velocity varying with the distance travelled according to the equation $$\displaystyle v= \alpha \sqrt{x},$$ where a is a constant. Find the total work done by all the forces during a displacement from $$x=0$$ to $$x=d$$.

A
$$\displaystyle \frac{1}{2}ma ^{2}b$$

B
$$\displaystyle \frac{1}{4}ma ^{2}b$$

C
$$ma ^{2}b$$

D
$$2\ ma ^{2}b$$

Solution

$$v=\alpha\sqrt{x}$$
$$\implies \dfrac{dx}{dt}=\alpha\sqrt{x}$$
$$\implies 2\sqrt{x}=\alpha t$$
$$\implies x=\dfrac{\alpha^2 t^2}{4}$$
Work done by forces$$=$$$$\int_0^b Fdx=_0^b\int madx$$
$$=_0^b\int m\dfrac{d^2x}{dt^2}dx$$
$$=_0^b\int m(\dfrac{2\alpha^2}{4})dx$$
$$=\dfrac{1}{2}m\alpha^2 b$$
Question 5
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Directions For Questions

A small particle of mass $$m$$ attached with a light inextensible thread of length $$L$$ is moving in a vertical circle. In the given case particle is moving in complete vertical circle and ratio of its maximum to minimum velocity is $$2:1$$.

...view full instructions

The kinetic energy of particle at the lower most position is

A
$$\displaystyle \frac{4mgL}{3}$$

B
$$2mgL$$

C
$$\displaystyle \frac{8mgL}{3}$$

D
$$\displaystyle \frac{2mgL}{3}$$

Solution

As $$v= 2\sqrt{\dfrac{gL}{3}}$$
Also, $$u= 2v \implies u= 4\sqrt{\dfrac{gL}{3}}$$
Thus $$K.E_ A= \dfrac{1}{2}mu^2= \dfrac{8mgL}{3}$$
Question 6
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A block of mass $$m=2$$ kg is pulled by a force $$F=40$$ N upwards through a height $$h=2 m$$. Find the work done on the block by the applied force F and its weight mg. $$\displaystyle \left ( g= 10 m/s^{2} \right )$$

A
$$\displaystyle W_F = 80 J;W_{mg}= -40 J$$

B
$$\displaystyle W_F = 80 J;W_{mg}= 40 J$$

C
$$\displaystyle W_F = -80 J;W_{mg}= -40 J$$

D
$$\displaystyle W_F = -80 J;W_{mg}= 40 J$$

Solution

Weight $$mg=(2)(10)=20$$ N

Work done by the applied force $$\displaystyle W_{F}= Fh\cos 0^{\circ}$$

As the angle between force and displacement is $$\displaystyle 0^{\circ}$$

or $$\displaystyle W_{F}= \left ( 40 \right )\left ( 2 \right )\left ( 1 \right )= 80 J$$

Similarly, work done by its weight

$$\displaystyle W_{mg}= \left ( mg \right )\left ( h \right )\cos 180^{\circ}$$

or $$\displaystyle W_{mg}= \left ( 20 \right )\left ( 2 \right )\left ( -1 \right )= -40 J$$
Question 7
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A force $$F$$ acting on a particle varies with the position $$x$$ as shown in figure. Find the work done by this force in displacing the particle from $$\displaystyle x= 0\ m\:to\:x= 2\ m.$$

A
$$-10J$$

B
$$10J$$

C
$$5J$$

D
$$-5J$$

Solution

From $$\displaystyle x= 0\ m\:to\:x= 2 \ m.$$, displacement of particle and force acting on the particle both are along positive x-direction. Therefore, work done is positive and given by the area under $$F-x$$ graph,
or $$\displaystyle W= \frac{1}{2}\left ( 2 \right )\left ( 10 \right )= 10J$$
Question 8
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A block is constrained to move along x-axis under a force F=-2x. Here, F is in newton and x in metre. Find the work done by this force when the block is displaced from x=2 m to x=-4 m.

A
-4 J

B
-8 J

C
-12 J

D
-16 J

Solution

Work done by a body in moving an object from $$x_1$$ to $$x_2$$
$$=\int_{x_1}^{x_2}Fdx$$
$$=\int_{x_1}^{x_2}(-2x)dx$$
$$=x_1^2-x_2^2$$
$$=-12J$$
Question 9
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A force $$F=(2+x)$$ acts on a particle in x-direction where $$F$$ is in newton and $$x$$ in meter. Find the work done by this force during a displacement from $$x=1.0$$ m to $$x=2.0\ m$$.

A
$$3.5 \ J$$

B
$$7 \ J$$

C
$$4.5 \ J$$

D
$$5.5 \ J$$

Solution

As the force is variable, we shall find the work done in a small displacement from $$x$$ to $$x+dx$$ and then integrate it to find the total work. The work done in this small displacement is

$$\displaystyle dW= F\:dx= \left ( 2+x \right )dx$$

Thus, $$\displaystyle W= \int_{1.0}^{2.0}dW= \int_{1.0}^{2.0}\left ( 2+x \right )dx$$

$$\displaystyle = \left [ 2x+\frac{x^{2}}{2} \right ]_{1.0}^{2.0}= 3.5J$$
Question 10
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A block is constrained to move along x-axis under a force $$\displaystyle F= \frac{4}{x^{2}}\left ( x\neq 0 \right ).$$ Here, F is in newton and x in metre. Find the work done by this force when the block is displaced from x=4 m to x=2 m.

A
-1 J

B
-2 J

C
-3 J

D
-4 J

Solution

Work done in moving an object from $$x_1$$ to $$x_2$$
$$=\int_{x_1}^{x_2}Fdx$$
$$=\int_{x_1}^{x_2}\dfrac{4}{x^2}dx$$
$$=4(\dfrac{1}{x_1}-\dfrac{1}{x_2})$$
$$=4(\dfrac{1}{4}-\dfrac{1}{2})$$
$$=-1J$$