Work Energy and Power Test

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Work Energy and Power Test - 14

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Weekly Quiz Competition

Question 1
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This question has statement 1 and statement 2. Of the four choices given after the statements, choose the one that best describes the two statements.
If two springs $$\mathrm{S}_{1}$$ and $$\mathrm{S}_{2}$$ of the force constants $$\mathrm{k}_{1}$$ and $$\mathrm{k}_{2}$$, respectively, are stretched by the same force, it is found that more work is done on spring $$\mathrm{S}_{1}$$ than on spring $$\mathrm{S}_{2}$$.
Statement-l : When stretched by the same amount, work done on $$\mathrm{S}_{1}$$, will be more than that of $$\mathrm{S}_{2}$$
Statement-2: $$\mathrm{k}_{1}<\mathrm{k}_{2}$$.

A
Statement-1 is false, statement-2 is true

B
Statement-1 is true, statement-2 is false

C
Statement-1 is true, statement-2 is true, statement 2 is the correct explanation of statement- 1

D
Statement-1 is true, statement-2 is true, statement-2 is not the correct explanation of Statement-1

Solution

$$k_1x_1=k_2x_2=F$$

$$\displaystyle W_1= \frac{1}{2}kx_1^2= \frac{F^2}{2k_1}$$

Similarly for $$\displaystyle W_2= \frac{F^2}{2k_2}$$

Since $$\displaystyle W \propto \frac{1}{k}$$

$$W_1.W_2$$ $$\Rightarrow k_1<k_2$$, Statement-2 is true.

From Statement-1, $$W_2>W_1$$, hence it is false.
Question 2
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A body of mass starts moving from rest along x-axis so that its velocity varies as $$v = a\sqrt {s}$$ where $$a$$ is a constant and $$s$$ is the distance covered by the body. The total work done by all the forces acting on the body in the first seconds after the start of the motion is:

A
$$\dfrac {1}{8} ma^{4}t^{2}$$

B
$$4\ ma^{4}t^{2}$$

C
$$8 ma^{4}t^{2}$$

D
$$\dfrac {1}{4}ma^{4}t^{2}$$

Solution

Velocity of body $$v = a \sqrt s$$
acceleration $$a' = \dfrac{dv}{dt} = \dfrac{a}{2 \sqrt s} \dfrac{ds}{dt} = \dfrac{a^2}{2}$$
displacement $$s' = \dfrac{1}{2} a' t^2 = \dfrac{1}{2} \dfrac{a^2}{2} t^2$$
Work done $$W = force \times displacement$$
$$W = ma' \times s' = \dfrac{1}{8} ma^4 t^2$$
Question 3
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When a rubber-band is stretched by a distance $$x$$, it exerts a restoring force of magnitude $$F = ax + bx^2$$ where $$a$$ and $$b$$ are constants. The work done in stretching the unstretched rubber band by $$L$$ is

A
$$\displaystyle \dfrac{aL^2}{2} + \dfrac{bL^3}{3}$$

B
$$\displaystyle \dfrac{1}{2} \left ( \dfrac{aL^2}{2} + \dfrac{bL^3}{3}\right )$$

C
$$aL^2 + bL^3$$

D
$$\displaystyle \dfrac{1}{2} (aL^2 + bL^3)$$

Solution

$$F = ax + bx^2$$

$$dw = Fdx$$

$$W = \displaystyle \int_0^L (ax + bx^2) dx$$

$$W = \displaystyle \frac{aL^2}{2} + \frac{bL^3}{3}$$
Question 4
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Directions For Questions

A small block of mass M moves on a frictionless surface of an inclined plane, as shown in figure. The angle of the incline suddenly changes from $$60^{o}$$ to $$30^{o}$$ at point B. The block is initially at rest at A. Assume that collisions between the block and the incline are totally inelastic $$(g = 10 m/s^{2}).$$

...view full instructions

If collision between the block and the incline is completely elastic, then the vertical (upward) component of the velocity of the block at point B, immediately after it strikes the second incline is

A
$$\sqrt{30}\mathrm{m}/\mathrm{s}$$

B
$$\sqrt{15}\mathrm{m}/\mathrm{s}$$

C
0

D
$$-\sqrt{15}\mathrm{m}/\mathrm{s}$$

Solution

After collision, vertical velocity will be
$$vsin({ 30 }^{ 0 })cos({ 30 }^{ 0 })-vcos({ 30 }^{ 0 })cos({ 60 }^{ 0 })=0$$.
Question 5
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What is the minimum velocity with which a body of mass $$m$$ must enter a vertical loop of radius $$R$$ so that it can complete the loop?

A
$$\sqrt{gR}$$

B
$$\sqrt{2gR}$$

C
$$\sqrt{3gR}$$

D
$$\sqrt{5gR}$$

Solution

To just complete the circle, at the highest point, tension is zero and the gravitational force provides the necessary centripetal force.
Hence, $$\dfrac{mv_{top}^2}{R}=mg.............(i)$$

Applying conservation of energy at the top and bottom points,
$$\dfrac{1}{2}mv_{bottom}^2=mg \times 2R + \dfrac{1}{2}mv_{top}^2.........(ii)$$

Substituting $$mv_{top}^2$$ from (i) in (ii) and solving,
$$v_{bottom}=\sqrt {5gR}$$
Question 6
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Two similar spring P and Q have spring constants $$K_P$$ and $$K_Q$$, such that $$K_P > K_Q$$. They are stretched, first by the same amount (case a), then by the same force (case b). The work done by the spring $$W_P$$ and $$W_Q$$ are related as, in case (a) and case (b), respectively:

A
$$W_P > W_Q; W_Q > W_P$$

B
$$W_P < W_Q; W_Q < W_P$$

C
$$W_P = W_Q; W_P > W_Q$$

D
$$W_P = W_Q; W_P = W_Q$$

Solution

Case a) both the springs are stretched by same amount, energy stored springs in P and Q will be

$$W_P=\dfrac{1}{2}K_P\Delta x^2$$ and $$W_Q=\dfrac{1}{2}K_Q\Delta x^2$$...(i)

given $$K_P>K_Q \Rightarrow W_P>W_Q$$...(ii)

Case b) both the springs are stretched by same force, energy stored springs in P and Q will be

$$W_P=\frac{1}{2}K_P\Delta x_P^2$$ and $$W_Q=\frac{1}{2}K_Q\Delta x_Q^2$$...(iii)

given $$K_P>K_Q \Rightarrow \Delta x_P<\Delta x_Q$$...(iv)
and $$ F=K_P\Delta x_P=K_Q\Delta x_Q$$...(v)

$$W_P=\dfrac{1}{2}K_P\Delta x_P \Delta x_P$$ and $$W_Q=\dfrac{1}{2}K_Q\Delta x_Q \Delta x_Q$$...(vi)

from equation (v) substituting $$F=K_P\Delta x_P=K_Q\Delta x_Q$$ in equation (vi)

$$W_P=\frac{1}{2}Fx_P$$ and $$W_Q=\dfrac{1}{2}F x_Q$$

given $$\Delta x_P>\Delta x_Q \Rightarrow W_P<W_Q$$...(vii)
hence, correct answer is option A.
Question 7
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A ball is suspended by a thread of length L at the point O on a wall which is inclined to the vertical by $$\alpha$$. The thread with the ball is displaced by a small angle $$\beta$$ away from the vertical and also away from the wall. If the ball is released, the period of oscillation of the pendulum when $$\beta > \alpha$$ will be

A
$$\displaystyle \sqrt{\frac{L}{g}}\left[\pi+2sin^{-1}\frac{\alpha}{\beta}\right]$$

B
$$\displaystyle \sqrt{\frac{L}{g}}\left[\pi-2sin^{-1}\frac{\alpha}{\beta}\right]$$

C
$$\displaystyle \sqrt{\frac{L}{g}}\left[2sin^{-1}\frac{\alpha}{\beta}-\pi\right]$$

D
$$\displaystyle \sqrt{\frac{L}{g}}\left[2sin^{-1}\frac{\alpha}{\beta}+\pi\right]$$

Solution

The motion of the string can be represented as an angular displacement from the mean position,given as: $$\theta=\beta sin(\omega t)$$
Time taken from displacement from mean position to $$\beta$$ and back to mean position=$$t_1=\dfrac{T}{2}$$
where $$T$$ is the time period of the oscillation without the wall barrier.
Thus $$t_1=\dfrac{1}{2}\times 2\pi\sqrt{\dfrac{L}{g}}=\pi\sqrt{\dfrac{L}{g}}$$
Time taken to displace from mean position to $$\alpha$$ can be found by:
$$\alpha=\beta sin(\omega t)$$
$$\implies t=\dfrac{1}{\omega}sin^{-1}\dfrac{\alpha}{\beta}$$
$$=\sqrt{\dfrac{L}{g}}sin^{-1}\dfrac{\alpha}{\beta}$$
Thus time taken to displace from mean position to $$\alpha$$ and back to mean position = $$t_2=2\sqrt{\dfrac{L}{g}}sin^{-1}\dfrac{\alpha}{\beta}$$
Thus the time period of the oscillation = $$t_1+t_2=\sqrt{\dfrac{L}{g}}[\pi+2sin^{-1}\dfrac{\alpha}{\beta}]$$
Question 8
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The force on a particle as the function of displacement x(in x-direction) is given by $$F=10+0.5x$$
The work done corresponding to displacement of particle from $$x=0$$ to $$x=2$$ unit is?

A
$$25$$ J

B
$$29$$ J

C
$$21$$ J

D
$$18$$ J

Solution

Given that $$F=10+0.5x$$
Work done corresponding to the particle from $$x=0$$ to $$x=2$$ unit is, $$W= \int_{0}^{2} F.dx= \int_{0}^{2} (10+0.5x)dx= [10x+ \dfrac{0.5x^2}{2}]_0^2= 20+1= 21 J$$
Question 9
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How much work must be done by a force on $$50\ $$$$kg$$ body in order to accelerate it from rest to $$20\ m/s$$ in $$10\ $$$$s$$ ?

A
$$10^3\ $$$$J$$

B
$$10^4\ $$$$J$$

C
$$2\times 10^3\ $$$$J$$

D
$$4\times 10^4\ $$$$J$$

Solution

Acceleration that causes the given rise in speed in given time = $$a=\dfrac{v}{t}=\dfrac{20\ m/s}{10\ s}=2\ m/s^2$$
Thus, force acting on the body = $$F=ma=100\ N$$
Distance travelled in the given time = $$s=\dfrac{1}{2}at^2=100\ m$$

Thus, work done in given time = $$F.s=10^4\ J$$
Question 10
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The amount of work done in stretching a spring from a stretched length of 10 cm to a stretched length of 20 cm is-

A
Equal to the work done in stretching it from 20 cm to 30 cm

B
Less than the work done in stretching it from 20 cm to 30 cm.

C
More than the work done in stretching it from 20 cm to 30 cm.

D
Equal to the work done in stretching it from 0 to 10 cm.

Solution

Work done in stretching a spring to x length is $$W=\dfrac{1}{2}kx^2$$
$$\therefore$$ Work one in stretching from $$x_1$$ to $$x_2$$ is proportional to $$W \propto x_2^2-x_1^2$$
hence $$\dfrac{W_{20,30}}{W_{10,20}}=\dfrac{500}{300}$$
so, 10 to 20 takes less work when compared to 20 to 30.

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Re-Attempt Weekly Quiz Competition

Work Energy and Power Test - 14

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Work Energy and Power Test - 14

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SHARING IS CARING

Question 1

Statement-1 is false, statement-2 is true

Statement-1 is true, statement-2 is false

Statement-1 is true, statement-2 is true, statement 2 is the correct explanation of statement- 1

Statement-1 is true, statement-2 is true, statement-2 is not the correct explanation of Statement-1

Solution

Question 2

$$\dfrac {1}{8} ma^{4}t^{2}$$

$$4\ ma^{4}t^{2}$$

$$8 ma^{4}t^{2}$$

$$\dfrac {1}{4}ma^{4}t^{2}$$

Solution

Question 3

$$\displaystyle \dfrac{aL^2}{2} + \dfrac{bL^3}{3}$$

$$\displaystyle \dfrac{1}{2} \left ( \dfrac{aL^2}{2} + \dfrac{bL^3}{3}\right )$$

$$aL^2 + bL^3$$

$$\displaystyle \dfrac{1}{2} (aL^2 + bL^3)$$

Solution

Question 4

$$\sqrt{30}\mathrm{m}/\mathrm{s}$$

$$\sqrt{15}\mathrm{m}/\mathrm{s}$$

0

$$-\sqrt{15}\mathrm{m}/\mathrm{s}$$

Solution

Question 5

$$\sqrt{gR}$$

$$\sqrt{2gR}$$

$$\sqrt{3gR}$$

$$\sqrt{5gR}$$

Solution

Question 6

$$W_P > W_Q; W_Q > W_P$$

$$W_P < W_Q; W_Q < W_P$$

$$W_P = W_Q; W_P > W_Q$$

$$W_P = W_Q; W_P = W_Q$$

Solution

Question 7

$$\displaystyle \sqrt{\frac{L}{g}}\left[\pi+2sin^{-1}\frac{\alpha}{\beta}\right]$$

$$\displaystyle \sqrt{\frac{L}{g}}\left[\pi-2sin^{-1}\frac{\alpha}{\beta}\right]$$

$$\displaystyle \sqrt{\frac{L}{g}}\left[2sin^{-1}\frac{\alpha}{\beta}-\pi\right]$$

$$\displaystyle \sqrt{\frac{L}{g}}\left[2sin^{-1}\frac{\alpha}{\beta}+\pi\right]$$

Solution

Question 8

$$25$$ J

$$29$$ J

$$21$$ J

$$18$$ J

Solution

Question 9

$$10^3\ $$$$J$$

$$10^4\ $$$$J$$

$$2\times 10^3\ $$$$J$$

$$4\times 10^4\ $$$$J$$

Solution

Question 10

Equal to the work done in stretching it from 20 cm to 30 cm

Less than the work done in stretching it from 20 cm to 30 cm.

More than the work done in stretching it from 20 cm to 30 cm.

Equal to the work done in stretching it from 0 to 10 cm.

Solution

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