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

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A body of mass m is moving in a circle of radius r with a constant speed v. The force on the body is $\frac{m v^{2}}{r}$ and is directed towards the centre. What is the work done by this force in moving the body over half the circumference of the circle

$\frac{m v^{2}}{π r^{2}}$

Zero

$\frac{m v^{2}}{r^{2}}$

$\frac{π r^{2}}{m v^{2}}$

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Explanation

Work done by centripetal force is always zero, because force and instantaneous displacement are always perpendicular.

$W = \vec{F} . \vec{s} = F s \cos θ = F s \cos (90 °) = 0$

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A man pushes a wall and fails to displace it. He does

Negative work

Positive but not maximum work

No work at all

Maximum work

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Explanation

No displacement is there.

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A body moves a distance of 10 m along a straight line under the action of a force of 5 N. If the work done is 25 joules, the angle which the force makes with the direction of motion of the body is

0°

30°

60°

90°

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Explanation

$W = F s \cos θ$

⇒ $\cos θ = \frac{W}{F s} = \frac{25}{50} = \frac{1}{2}$

⇒ $θ = 60 °$

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A force acts on a 30 gm particle in such a way that the position of the particle as a function of time is given by $x = 3 t - 4 t^{2} + t^{3}$ , where x is in metres and t is in seconds. The work done during the first 4 seconds is

5.28 J

450 mJ

490 mJ

530 mJ

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Explanation

$v = \frac{d x}{d t} = 3 - 8 t + 3 t^{2}$

∴ $v_{0} = 3 m / s$ and $v_{4} = 19 m / s$

$W = \frac{1}{2} m (v_{4}^{2} - v_{0}^{2})$ (According to work energy theorem)

$= \frac{1}{2} \times 0 .03 \times (19^{2} - 3^{2}) = 5 .28 J$

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The work done in pulling up a block of wood weighing 2 kN for a length of 10m on a smooth plane inclined at an angle of 15° with the horizontal is (given: sin15°=0.2588):

4.36 kJ

5.17 kJ

8.91 kJ

9.82 kJ

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A force $\vec{F} = 5 \hat{i} + 6 \hat{j} - 4 \hat{k}$ acting on a body, produces a displacement $\vec{s} = 6 \vec{i} + 5 \vec{k} .$ Work done by the force is

18 units

15 units

12 units

10 units

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Explanation

$W = \vec{F} . \vec{s} = (5 \hat{i} + 6 \hat{j} - 4 \hat{k}) . (6 \hat{i} + 5 \hat{k}) = 30 - 20 = 10$ units

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A body of mass 6kg is under a force which causes displacement in it given by $S = \frac{t^{2}}{4}$ metres where t is time. The work done by the force in 2 seconds is-

12 J

9 J

6 J

3 J

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Explanation

$s = \frac{t^{2}}{4}$ ∴ $d s = \frac{t}{2} d t$

$F = m a = \frac{m d^{2} s}{d t^{2}} = \frac{6 d^{2}}{d t^{2}} [\frac{t^{2}}{4}] = 3 N$

Now $W = \int_{0}^{2} F d s = \int_{0}^{2} 3 \frac{t}{2} d t = \frac{3}{2} {[\frac{t^{2}}{2}]}_{0}^{2} = \frac{3}{4} [{(2)}^{2} - {(0)}^{2}] = 3 J$

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A force of $(3 \hat{i} + 4 \hat{j})$ Newton acts on a body and displaces it by $(3 \hat{i} + 4 \hat{j}) m .$ The work done by the force is

10 J

12 J

16 J

25 J

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Explanation

$W = \vec{F} . \vec{s}$

$= (3 \hat{i} + 4 \hat{j}) . (3 \hat{i} + 4 \hat{j}) = 9 + 16 = 25 J$

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A force $\vec{F} = 6 \hat{i} + 2 \hat{j} - 3 \hat{k}$ acts on a particle and produces a displacement of $\vec{s} = 2 \hat{i} - 3 \hat{j} + x \hat{k} .$ If the work done is zero, the value of x is

– 2

1/2

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Explanation

$W = \vec{F} . \vec{s} = (6 \hat{i} + 2 \hat{j} - 3 \hat{k}) . (2 \hat{i} - 3 \hat{j} + x \hat{k}) = 0$

$12 - 6 - 3 x = 0$ ⇒ $x = 2$

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A particle moves from position ${\vec{r}}_{1} = 3 \hat{i} + 2 \hat{j} - 6 \hat{k}$ to position ${\vec{r}}_{2} = 14 \hat{i} + 13 \hat{j} + 9 \hat{k}$ under the action of force $4 \hat{i} + \hat{j} + 3 \hat{k} N .$ The work done will be

100 J

50 J

200 J

75 J

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Explanation

Work done = force x displacement

$W = \vec{F} . (\vec{r_{2}} - \vec{r_{1}}) = (4 \hat{i} + \hat{j} + 3 \hat{k}) (11 \hat{i} + 11 \hat{j} + 15 \hat{k})$

$W = 44 + 11 + 45 = 100 J o u l e$

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