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Work Energy And Power NEET Questions — Free MCQ Practice

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A constant force F is applied on a body. The power (P) generated is related to the time elapsed (t) as

[This question is only for Dropper and XII batch]

A.

$P \propto t^{2}$

B.

$P \propto t$

C.

$P \propto \sqrt{t}$

D.

$P \propto t^{3 / 2}$

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Explanation

$F = \frac{m d v}{d t} ∴ F d t = m d v \Rightarrow v = \frac{F}{m} t$

Now $P = F \times v = F \times \frac{F}{m} t = \frac{F^{2} t}{m}$

If force and mass are constants then $P \propto t$

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A spring 40 mm long is stretched by the application of a force. If 10 N force required to stretch the spring through 1 mm, then work done in stretching the spring through 40 mm is

A.

84J

B.

68J

C.

23J

D.

8J

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Explanation

Here

$k = \frac{F}{x} = \frac{10}{1 \times 10^{- 3}} = 10^{4} N / m W = \frac{1}{2} k x^{2} = \frac{1}{2} \times 10^{4} \times {(40 \times 10^{- 3})}^{2} = 8 J$

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Two springs with spring constants $k_{1}$ = 1500 N/m and $k_{2}$ = 3000 N/m are stretched by the same force. The ratio of potential energy stored in the springs will be

A.

2:1

B.

1:2

C.

4:1

D.

1:4

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Explanation

Since force is same

$F = k_{1} x_{1} = k_{2} x_{2} So, \frac{x_{1}}{x_{2}} = \frac{k_{2}}{k_{1}} Also, \frac{U_{1}}{U_{2}} = \frac{\frac{1}{2} k_{1} {x^{2}}_{1}}{\frac{1}{2} k_{2} {x^{2}}_{2}} = \frac{k_{2}}{k_{1}} = \frac{3000}{1500} = \frac{2}{1}$

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A particle of mass 10 kg is moving with velocity of $10 \sqrt{x}$ m/s, where x is displacement . The work done by net force during the displacement of particle form x = 4 to x = 9 m is

A.

1250 J

B.

1000 J

C.

3500 J

D.

2500 J

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Explanation

Using work-energy theorem

$W_{A l l f o r c e} = {(K E)}_{f} - {(K E)}_{i} = \frac{1}{2} \times 10 \times {(30)}^{2} - \frac{1}{2} \times 10 \times {(20)}^{2} = 5 [50 \times 10] = 2500 J$

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A body starts moving from rest in straight line under a constant power source. Its displacement in time t is proportional to

A.

$t^{1 / 2}$

B.

t

C.

$t^{3 / 2}$

D.

$t^{2}$

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Explanation

$P = F . v P = m a . v P = m v \frac{d v}{d t} P d t = m v d v P \int d t = m \int v d v P . t = m \frac{v^{2}}{2} v = \sqrt{\frac{2 P t}{m}} d x = \sqrt{\frac{2 P t}{m}} d t \int d x = \sqrt{\frac{2 P}{m}} \int \sqrt{t} d t x = \sqrt{\frac{2 P}{m}} \times \frac{2}{3} t^{3 / 2}$

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The relation between velocity (v) and time (t) is $vα \sqrt{t}$ , then which one of the following quantity is constant:

A.

Force

B.

Power

C.

Momentum

D.

Kinetic Energy

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Explanation

2.

$v = k \sqrt{t}$ {where k is proportionality constant}

$a = \frac{dv}{dt} = \frac{k}{2 \sqrt{t}}$

velocity, acceleration be the time-dependent, so K.E. force and momentum is also time-varying

$Power = force \times velocity = \frac{mk}{2 \sqrt{t}} k \sqrt{t} = \frac{{mk}^{2}}{2}$

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A steel wire can withstand a load up to 2940 N. A load of 150 kg is suspended from a rigid support. The maximum angle through which the wire can be displaced from the mean position, so that the wire does not break when the load passes through the position of equilibrium, is (2008 E)

A.

30 $°$

B.

60 $°$

C.

80 $°$

D.

85 $°$

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A body is thrown vertically up with certain initial velocity, the potential and kinetic energies of the body are equal at a point P in its path. If the same body is thrown with double the velocity upwards, the ratio of potential and kinetic energies of the body when it crosses the same point, is

A.

1:1

B.

1:4

C.

1:7

D.

1:8

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Explanation

$W h e n t h e b o d y i s t h r o w n w i t h v e l o c i t y = v; T E = K E_{i n i t i a l} = \frac{1}{2} m {(v)}^{2} = K A t t h e p o i n t, w h e n K E a n d P E a r e e q u a l : K E + P E = T E \Rightarrow K E = P E = \frac{K}{2} W h e n t h e b o d y i s t h r o w n w i t h v e l o c i t y = 2 v; T E' = K E'_{i n i t i a l} = \frac{1}{2} m {(2 v)}^{2} = 4 \times \frac{1}{2} m {(v)}^{2} = 4 K P E a t t h e s a m e p o i n t w i l l b e s a m e = \frac{K}{2} K E a t t h a t p o i n t = 4 K - \frac{K}{2} = \frac{7 K}{2} R a t i o o f P E t o K E = 1 : 7$

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A body is displaced from (0,0) to (1m,1m) along the path x=y by a force $F = (x^{2} \hat{j} + y \hat{i}) N$ . The work done by this force will be :

A.

$\frac{4}{3} J$

B.

$\frac{5}{6} J$

C.

$\frac{3}{2} J$

D.

$\frac{7}{5} J$

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Explanation

$dW = \vec{F} . \vec{dr} = (y \hat{i} + x^{2} \hat{j}) . (dx \hat{i} + dy \hat{j}) = ydx + x^{2} dy From y = x, putting y in terms of x and x in terms of y in above relations dW = xdx + y^{2} dy Integrating both sides - W = \int_{0}^{1} xdx + \int_{0}^{1} y^{2} dy = {[\frac{x^{2}}{2}]}_{0}^{1} + {[\frac{y^{3}}{3}]}_{0}^{1} = \frac{1}{2} + \frac{1}{3} = \frac{5}{6} J$

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A force F is applied on a body which moves with a velocity v in the direction of the force, then the power will be

A.

${Fv}^{2}$

B.

Fv

C.

$F / v^{2}$

D.

F/v

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Explanation

$(2) P = Fv = \frac{W}{t} = F (\frac{s}{t})$

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

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