Work Energy and Power Test

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Work Energy and Power Test - 28

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Question 1
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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 $$2:1$$.

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Minimum velocity of the particle is

A
$$\displaystyle 4\sqrt{\frac{gL}{3}}$$

B
$$\displaystyle 2\sqrt{\frac{gL}{3}}$$

C
$$\displaystyle \sqrt{\frac{gL}{3}}$$

D
$$\displaystyle 3\sqrt{\frac{gL}{3}}$$

Solution

The maximum and minimum velocities will be at A and B respectively.
Work-energy theorem, $$-mg(2L)= \dfrac{1}{2}mv^2-\dfrac{1}{2}mu^2$$
Given: $$u:v= 2:1$$
$$-mg(2L)= \dfrac{1}{2}mv^2-\dfrac{1}{2}m(2v)^2$$
$$\implies v= 2\sqrt{\dfrac{gL}{3}}$$
Question 2
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Directions For Questions

Bob B of the pendulum AB is given an initial velocity $$\displaystyle \sqrt{3Lg}$$ in horizontal direction. Find the maximum height of the bob from the starting point:

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if AB is a massless string.

A
$$\displaystyle \frac{10L}{27}$$

B
$$\displaystyle \frac{20L}{27}$$

C
$$\displaystyle \frac{30L}{27}$$

D
$$\displaystyle \frac{40L}{27}$$

Solution

When the string has traversed an angle $$\theta$$ , then

$$T-mgcos\theta =\dfrac { m{ v }^{ 2 } }{ l }$$ , and conservation of energy gives

$$\dfrac { m{ u }^{ 2 } }{ 2 } =\dfrac { m{ v }^{ 2 } }{ 2 } +mgl(1-cos\theta ).$$ then the string becomes slacked tension in the string becomes zero. Solving the above two equations to get the value of $$\theta$$

$$\Rightarrow cos\theta =\dfrac { -1 }{ 3 } and { v }^{ 2 }={ u }^{ 2 }-2gl(1-cos\theta )=\dfrac { gl }{ 3 } $$ from that point the maximum height reached is $$\dfrac { { v }^{ 2 }{ sin }^{ 2 }\left( \pi -\theta \right) }{ 2g }$$ maximum height $$= l(1-cos\theta )+\dfrac { { v }^{ 2 }{ sin }^{ 2 }\left( \pi -\theta \right) }{ 2g } =\dfrac { 4l }{ 3 } +\dfrac { 4l }{ 27 } =\dfrac { 40l }{ 27 } $$
Question 3
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When an arrow is released from a bow, potential energy changes into kinetic energy.

A
True

B
False

C
Ambiguous

D
Data insufficient

Solution

When an arrow is drawn back by a bow, the work done by us in stretching the bowstring gets stored at potential energy in the bow. This potential energy of bow is transformed into kinetic energy when the bowstring is released and this gives kinetic energy to the arrow.
Question 4
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An object in a given position can have a certain potential energy with respect to one level and a different value of potential energy with respect to another level.

A
True

B
False

C
Ambiguous

D
Data insufficient

Solution

Potential energy of an object always defined with respect to reference point. If location of reference point changes, there is a possibility that potential at that point would change.
In general, we write the potential energy of an object at a height $$h$$ from earth surface $$U=mgh$$. Here we take earth surface as a zero potential level or reference point.
Hence, Given statement is true. Option A
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

Velocity of particle when it is moving vertically downward is

A
$$\displaystyle \sqrt{\frac{10gL}{3}}$$

B
$$\displaystyle 2\sqrt{\frac{gL}{3}}$$

C
$$\displaystyle \sqrt{\frac{8gL}{3}}$$

D
$$\displaystyle \sqrt{\frac{13gL}{3}}$$

Solution

Work-energy theorem for A to C, $$-mgL= \dfrac{1}{2}m(v')^2- \dfrac{1}{2}mu^2$$ where $$u= 4\sqrt{\dfrac{gL}{3}}$$
$$-mgL= \dfrac{1}{2}m(v')^2- \dfrac{8mgL}{3}$$
$$\implies v'= \sqrt{\dfrac{10 gL}{3}}$$
Question 6
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A boy is whirling a stone tied at one end such that the stone is in uniform circular motion.Which of the following statement is correct?

A
The velocity of stone at A is equal to the velocity of stone B

B
The speed of stone at A is equal to the speed of stone at B.

C
The centripetal force is radially outward in the string.

D
All of the above statements are correct.

Solution

In uniform circular motion speed is constant while velocity being a vector quantity is constantly changing as its direction keeps changing. Centripetal force acts inwards towards the center to counterbalance the centrifugal force acting outwards from the center.
Question 7
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A car is accelerated on a levelled road and attains a velocity 4 times of its initial velocity. In this process, the potential energy of the car

A
does not change

B
becomes twice of initial

C
becomes 4 times of initial

D
becomes 16 times of initial

Solution

Answer is A.

The potential energy is the energy that an object has due to its position in a force field or that a system has due to the configuration of its parts.
The potential energy of the car remains the same and will not change as the road is leveled and the height of the body remains the same, although its speed increases.
Question 8
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A mass is performing vertical circular motion (see figure). If the average velocity of the particle is increased, then at which point is the maximum breaking possibility of the string:

A
A

B
B

C
C

D
D

Solution

Tension at any point in vertical motion is given by :
$$T=\dfrac{mv^{2}}{l}+mg\cos\theta$$
where $$\theta=$$ angular displacement from lowest point ,
$$l=$$ length of string
$$m=$$ mass of string
It is clear that tension at the lowest point ($$B$$) is greatest than at other points ($$A,C,D$$). If we increase average velocity, tension will increase at lowest point, therefore at point B, string has maximum possibility of break.
Question 9
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A stone of mass $$1\;kg$$ is tied to the end of a string of $$1\;m$$ long. It is whirled in a vertical circle. If the velocity of the stone at the top be $$4\;m/s$$. What is the tension in the string?

A
$$\;6\;N$$

B
$$\;16\;N$$

C
$$\;5\;N$$

D
$$\;10\;N$$

Solution

Given : $$v =4$$ $$m/s$$ $$m =1 kg$$ $$r = 1$$ m
Using circular motion equation : $$\dfrac{mv^2}{r} =T + mg$$
$$\implies$$ $$ T =\dfrac{mv^2}{r} -mg $$

$$\therefore$$ $$ T =\dfrac{1 \times 4^2}{1} -1 \times 10 $$ $$\implies T = 6 $$ $$N$$
Question 10
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Two weights of 5 kg and 10 kg are placed on a horizontal table of height 1.5 m. Which will have more potential energy?

A
5 kg

B
10 kg

C
Both will have equal energy

D
None of the above

Solution

We know that , P.E $$ = mgh $$
So, It is directly proportional to height and mass.
Since both the weights are at the same height, so the weight with a larger mass will have more potential energy.
Since $$10$$kg object has a larger mass than a $$5$$ kg,
so, the potential energy of a $$10$$kg mass will be greater.