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Work Energy And Power NEET Questions — Free MCQ Practice

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Three different objects of masses $m_{1}, m_{2}$ and m₃ are allowed to fall from rest and from the same point ‘O’ along three different frictionless paths. The speeds of the three objects, on reaching the ground, will be in the ratio of

A.

$m_{1} : m_{2} : m_{3}$

B.

$m_{1} : 2 m_{2} : 3 m_{3}$

C.

1 : 1 : 1

D.

$\frac{1}{m_{1}} : \frac{1}{m_{2}} : \frac{1}{m_{3}}$

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Explanation

Speed of the object at reaching the ground $v = \sqrt{2 g h}$

If heights are equal then velocity will also be equal.

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When a body moves with a constant speed along a circle

A.

No work is done on it

B.

No acceleration is produced in the body

C.

No force acts on the body

D.

Its velocity remains constant

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Explanation

When speed is constant in circular motion, it means work done by centripetal force is zero.

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A sphere of mass m is tied to end of a string of length l and rotated through the other end along a horizontal circular path with speed v. The work done by centripetal force in full horizontal circle is

A.

0

B.

$(\frac{m v^{2}}{l}) . 2 π l$

C.

$mg . 2πl$

D.

$(\frac{m v^{2}}{l}) . (l)$

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Explanation

Work done by centripetal force in uniform circular motion is always equal to zero.

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A ball is suspended by a thread of length l. What minimum horizontal velocity has to be imparted to the ball for it to reach the height of the suspension:

A.

gl

B.

2 gl

C.

$\sqrt{g l}$

D.

$\sqrt{2 g l}$

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Explanation

To reach the height of suspension l, the particle must have a vertical component of velocity âˆš(2gl) at the highest point. This vertical velocity can be obtained by imparting a horizontal velocity âˆš(2gl) at the lowest point, as the total velocity is âˆš(2) times the horizontal component.

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A body of mass m hangs at one end of a string of length l, the other end of which is fixed. It is given a horizontal velocity so that the string would just reach where it makes an angle of 60° with the vertical. The tension in the string at mean position is

A.

2 mg

B.

mg

C.

3 mg

D.

$\sqrt{3} m g$

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The kinetic energy k of a particle moving along a circle of radius R depends on the distance covered s as k = as² where a is a constant. The force acting on the particle is

A.

$2 a \frac{s^{2}}{R}$

B.

$2 a s {(1 + \frac{s^{2}}{R^{2}})}^{1 / 2}$

C.

2 as

D.

$2 a \frac{R^{2}}{s}$

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Explanation

According to the given problem, $\frac{1}{2} m v^{2} = a s^{2} \Rightarrow v = s \sqrt{\frac{2 a}{m}}$

So, $a_{R} = \frac{v^{2}}{R} = \frac{2 a s^{2}}{m R}$

Furthermore, as $a_{t} = \frac{d v}{d t} = \frac{d v}{d s} \cdot \frac{d s}{d t} = v \frac{d v}{d s}$

$a_{t} = [s \sqrt{\frac{2 a}{m}}] [\sqrt{\frac{2 a}{m}}] = \frac{2 a s}{m}$

So, $a = \sqrt{a_{R}^{2} + a_{t}^{2}} = \sqrt{{[\frac{2 a s^{2}}{m R}]}^{2} + {[\frac{2 a s}{m}]}^{2}}$

Hence $a = \frac{2 a s}{m} \sqrt{1 + {[s / R]}^{2}}$

∴ $F = m a = 2 a s \sqrt{1 + {[s / R]}^{2}}$

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A stone of mass 1 kg tied to a light inextensible string of length $L = \frac{10}{3} m$ is whirling in a circular path of radius L in a vertical plane. If the ratio of the maximum tension in the string to the minimum tension in the string is 4 and if g is taken to be 10 m/sec², the speed of the stone at the highest point of the circle is

A.

20 m/sec

B.

$10 \sqrt{3} m / \sec$

C.

$5 \sqrt{2} m / \sec$

D.

10 m/sec

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Explanation

Since the maximum tension T_B in the string moving in the vertical circle is at the bottom and minimum tension T_T is at the top.

∴ $T_{B} = \frac{m {v_{B}}^{2}}{L} + m g$ and $T_{T} = \frac{m v_{T}^{2}}{L} - m g$

∴ $\frac{T_{B}}{T_{T}} = \frac{\frac{m v_{B}^{2}}{L} + m g}{\frac{m v_{T}^{2}}{L} - m g} = \frac{4}{1}$ or $\frac{v_{B}^{2} + g L}{v_{T}^{2} - g L} = \frac{4}{1}$

or ${v_{B}}^{2} + g L = 4 v_{T}^{2} - 4 g L$ (1)

$∵ By energy conservation, \frac{1}{2} {mv}^{2}_{B} + 0 = mg (2 L) + \frac{1}{2} {mv}^{2}_{T} \Rightarrow v_{B}^{2} = v_{T}^{2} + 4 g L$

Put it in equation (1):

∴ $v_{T}^{2} + 4 g L + g L = 4 v_{T}^{2} - 4 g L$ ⇒ $3 v_{T}^{2} = 9 g L$

∴ $v_{T}^{2} = 3 \times g \times L = 3 \times 10 \times \frac{10}{3}$ or v_T = 10 m/sec

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A stone tied to a string of length L is whirled in a vertical circle with the other end of the string at the centre. At a certain instant of time, the stone is at its lowest position and has a speed u. The magnitude of the change in its velocity as it reaches a position where the string is horizontal is:

A.

$\sqrt{u^{2} - 2 g L}$

B.

$\sqrt{2 g L}$

C.

$\sqrt{u^{2} - g l}$

D.

$\sqrt{2 (u^{2} - g L)}$

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Explanation

Using conservation of energy :

$\frac{1}{2} m u^{2} - \frac{1}{2} m v^{2} = m g L$

⇒ $v = \sqrt{u^{2} - 2 g L}$

$| \vec{v} - \vec{u} | = \sqrt{u^{2} + v^{2}} = \sqrt{u^{2} + u^{2} - 2 g L} = \sqrt{2 (u^{2} - g L)}$

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Work done by a frictional force is

A.

Negative

B.

Positive

C.

Zero

D.

All of the above

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Explanation

Work done by friction can be positive, negative and zero depending upon the situation.

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A block of mass 50 kg slides over a horizontal distance of 1 m. If the coefficient of friction between their surfaces is 0.2, then work done against friction is

A.

98 J

B.

72J

C.

56 J

D.

34 J

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Explanation

$W = μ m g S = 0 .2 \times 50 \times 9 .8 \times 1 = 98 J$

Work Energy And Power MCQs for NEET — Physics Questions with Answers

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