Work Energy and Power Test

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Work Energy and Power Test - 36

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Weekly Quiz Competition

Question 1
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Which one of the following energies cannot be possessed by a body at rest?

A
Potential energy

B
Kinetic energy

C
Thermal energy

D
Magnetic energy

Solution

Kinetic energy is possessed by a body by virtue of its state of motion. So, a body at rest cannot possess kinetic energy. A body at rest will possess potential energy. Thermal and magnetic energies are irrespective of state of rest or of motion of a body.
Question 2
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A ball of mass $$m$$ is attached with the string of length $$R$$, rotating in circular motion, with instantaneous velocity $$v$$ and centripetal acceleration $$a$$.
Calculate the centripetal acceleration of the ball if the length of the string is doubled?

A
$${a}/{4}$$

B
$${a}/{2}$$

C
$$a$$

D
$$2a$$

E
$$4a$$

Solution

Centripetal acceleration $$a = \dfrac{v^2}{R}$$
Now the length of the string is doubled keeping the instantaneous velocity constant i.e $$R' = 2R$$
$$\therefore$$ New centripetal acceleration $$a' = \dfrac{v^2}{R'} = \dfrac{v^2}{2R} =\dfrac{a}{2}$$
Question 3
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A car standing at the top of a hill would be an example of what type of energy?

A
Kinetic

B
Topical

C
Potential

D
Beneficial

E
Residual

Solution

The car standing at the top of hill has attained a height above the ground level. Thus it has attained a gravitational potential energy given by $$mgh$$
where $$m$$ is the mass of the car,
$$g$$ is the gravitational constant,
$$h$$ is the height above ground that it has attained.
This energy can convert into the kinetic energy of the car if it starts coming down the hill, that is lose its potential energy.
Question 4
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A stone ties to a rope is rotated in a vertical circle with uniform speed. If the difference between the maximum and minimum tension in the rope is $$20\ N$$, mass of the stone in $$kg$$ is:
($$\displaystyle g=10{ ms }^{ -2 }$$)

A
$$1.0$$

B
$$1.5$$

C
$$0.5$$

D
$$0.75$$

Solution

Let the mass of the stone be $$m$$, and the angular speed with which it rotates be $$\omega$$.
Thus at the bottom most point,
$$T_b-mg=m\omega^2 l$$
And at the top most point,
$$T_t+mg=m\omega ^2 l$$
$$\implies T_b-T_t=2mg=20$$
$$\implies mg=10$$
$$\implies m=1\ kg$$
Question 5
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In vertical circular motion, the ratio of kinetic energy to potential energy at the horizontal position is ____________.

A
$$5:2$$

B
$$2:1$$

C
$$3:2$$

D
$$2:3$$

Solution

We assume that the object in motion just manages to complete the vertical circle. In that case,
$$v_{top} = \sqrt{gR}$$
Total energy at the topmost point $$=\dfrac12mgR + 2mgR = \dfrac52mgR$$
At the horizontal position $$PE = mgR$$
Total energy $$= \dfrac52mgR$$
$$\therefore KE = (\dfrac52-1) = \dfrac32mgR$$

$$\therefore \dfrac{KE}{PE} = \dfrac32$$
Question 6
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An object of mass $$40\ kg$$ is raised to a height of $$5\ m$$ above ground. If the object is allowed to fall down find its kinetic energy when it is half-way down.

A
1960 J

B
980 J

C
990 J

D
1000 J

Solution

Gravitational Potential energy, $$W = mgh$$
$$h=$$ vertical displacement $$=5\ m$$
$$m =$$ Mass of object $$= 40\ Kg$$
$$g =$$ Acceleration due to gravity $$= 9.8\ ms^{-2}$$
$$Kinetic $$ $$ Energy$$ = $$Loss$$ $$in$$ $$PE$$= $$40 \times 5 \times 9.8= 1960\ J$$
For half way down$$=\dfrac{1960}{2}= 980\ J$$
At this point i.e. half way object has an equal amount of potential and kinetic energy.
Question 7
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Directions For Questions

As shown above, a ball is attached to the string and it is swinging in vertical circle. The three position of ball are given as I, II and III on the circle.

...view full instructions

If the string were to break at point II, what would be the path of the ball?

A

B

C

D

E
None of these

Solution

When the bob is swinging in vertical circle the centripetal force and the pseudo force varies. Centripetal force causes the tension in the string.
Let centripetal force be $$C$$, pseudo force be $$P$$ and weight of bob be $$W$$.
At position 2, $$C = W$$
If the string were to break at this position, the only external force acting on the bob is $$W$$ (downward). Hence the bob would move vertically down.

Therefore option D is correct.
Question 8
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A $$50 kg$$ acrobat is swinging on a rope with a length of $$15 m$$ from one horizontal platform to another. Both platforms are at equal height.
If the maximum tension the rope can support is $$1200 N$$, which of the following answer best represents the maximum velocity the acrobat can reach without breaking the rope?

A
$$25 {m}/{s}$$

B
$$19 {m}/{s}$$

C
$$20 {m}/{s}$$

D
$$15 {m}/{s}$$

E
$$5 {m}/{s}$$

Solution

Given : $$T =1200$$ N $$m = 50$$ kg $$L =15$$ m
Let the maximum velocity of the acrobat be $$v$$.
From figure, $$\dfrac{mv^2}{L} =T - mg$$

$$\therefore$$ $$\dfrac{50 \times v^2}{15} =1200 - 50 \times 9.8$$ $$\implies v =\sqrt{213} \approx 15 $$ $$m/s$$
Question 9
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Directions For Questions

As shown above, a ball is attached to the string and it is swinging in vertical circle. The three position of ball are given as I, II and III on the circle.

...view full instructions

Which of the following statements is true?

A
The tension in the string is greater at points III than at point I

B
The tension in the string is greater at points I than at point III

C
The tension at point I is equal to the tension at point III

D
The tension in the string is greatest at point II

E
The tension in the string is same at points I, II and III

Solution

When the bob is swinging in vertical circle the centripetal force and the pseudo force varies. Centripetal force causes the tension in the string.
Let centripetal force be $$C$$, pseudo force be $$P$$ and weight of bob be $$W$$.
At position 1, $$C = P - W$$
At position 2, $$C = W$$
At position 3, $$C = P + W$$

As C at position 3 is greatest, the tension in the string is greatest at this position.
Therefore option A is correct.
Question 10
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Fifteen joules of work is done on object A, which is attached to an uncompressed spring, so that all the work done goes into compressing the spring.
Sixty joules of work is done on object B, which is attached to an uncompressed spring that is identical to the spring to which object A is attached; all the work done goes into compressing the spring.
How does the compression of the spring for object B compare to the compression of the spring for object A after this work is done?

A
The compression of the spring for object B is four times as much as it is for object A

B
The compression of the spring for object B is sixteen times as much as it is for object A

C
The compression of the spring for object B is two times as much as it is for object A

D
The compression of the spring for object B is less than two times as much as it is for object A (but not the same amount)

E
The compression of the spring for object B is the same as it is for object A

Solution

The work done in compressing the spring is stored as the potential energy of spring , given by
$$U=1/2kx^{2}$$
where $$x=$$ compression of spring ,
$$k=$$ spring constant ,
for object A , potential energy =work done
$$1/2kx_{A}^{2}=15$$ ...................................eq1
for object B , potential energy =work done
$$1/2kx_{B}^{2}=60$$ ....................................eq2
dividing eq2 by eq1 ,
$$ x_{B}^{2}/x_{A}^{2}=60/15=4$$
or $$x_{B}/x_{A}=2$$
or $$x_{B}=2x_{A}$$