JEE Main Work, Energy and Power PYQ

6. (JEE Main 2021 (Online) 1st September Evening Shift )

A body of mass 'm' dropped from a height 'h' reaches the ground with a speed of 0.8 $\sqrt{g h}$ . The value of workdone by the air-friction is :

A. $-$ 0.68 mgh

B. mgh

C. 1.64 mgh

D. 0.64 mgh

Correct Option is (A)

Given, the mass of the body = m

The height from which the body dropped = h

The speed of the body when reached the ground, $v_{f} = 0.8 \sqrt{g h}$

Initial velocity of the body, v = 0 m/s

Using the work-energy theorem,

Work done by gravity + Work done by air-friction = Final kinetic energy $-$ Initial kinetic energy.

$W_{m g} + W_{a i r - f r i c t i o n} = \frac{1}{2} m v_{f}^{2} - \frac{1}{2} m v_{i}^{2}$

Here, work done by gravity = mgh

$\Rightarrow m g h + W_{a i r - f r i c t i o n} = \frac{1}{2} m (0.8 \sqrt{g h})^{2} - \frac{1}{2} m (0)^{2}$

$\Rightarrow W_{a i r - f r i c t i o n} = \frac{0.64 m g h}{2} - m g h$

$\Rightarrow 0.32 m g h - m g h = - 0.68 m g h$

The value of the work done by the air friction is $-$ 0.68 mgh.

7. (JEE Main 2022 (Online) 27th July Morning Shift )

Sand is being dropped from a stationary dropper at a rate of $0.5 {kgs}^{- 1}$ on a conveyor belt moving with a velocity of $5 {ms}^{- 1}$ . The power needed to keep the belt moving with the same velocity will be :

A. 1.25 W

B. 2.5 W

C. 6.25 W

D. 12.5 W

Correct Option is (D)

$\frac{d m}{d t} = 0.5$ kg/s

$v = 5$ m/s

$F = \frac{v d m}{d t} = 2.5$ kg m/s²

$P = \overset{―}{F} . \overset{―}{v} = (2.5) (5)$ W

$= 12.5$ W

8. (JEE Main 2021 (Online) 27th July Evening Shift )

An automobile of mass 'm' accelerates starting from origin and initially at rest, while the engine supplies constant power P. The position is given as a function of time by :

A. ${(\frac{9 P}{8 m})}^{\frac{1}{2}} t^{\frac{3}{2}}$

B. ${(\frac{8 P}{9 m})}^{\frac{1}{2}} t^{\frac{2}{3}}$

C. ${(\frac{9 m}{8 P})}^{\frac{1}{2}} t^{\frac{3}{2}}$

D. ${(\frac{8 P}{9 m})}^{\frac{1}{2}} t^{\frac{3}{2}}$

Correct Option is (D)

P = const.

$P = F v = \frac{m v^{2} d v}{d x}$

$\int_{0}^{x} \frac{P}{m} d x = \int_{0}^{v} v^{2} d v$

$\frac{P x}{m} = \frac{v^{3}}{3}$

${(\frac{3 P x}{m})}^{1 / 3} = v = \frac{d x}{d t}$

${(\frac{3 P}{m})}^{1 / 3} \int_{0}^{t} d t = \int_{0}^{x} x^{- 1 / 3} d x$

$\Rightarrow x = {(\frac{8 P}{9 m})}^{1 / 2} t^{3 / 2}$

9. (JEE Main 2021 (Online) 20th July Evening Shift )

A body at rest is moved along a horizontal straight line by a machine delivering a constant power. The distance moved by the body in time 't' is proportional to :

A. $t^{\frac{3}{2}}$

B. $t^{\frac{1}{2}}$

C. $t^{\frac{1}{4}}$

D. $t^{\frac{3}{4}}$

Correct Option is (A)

P = constant

$\frac{1}{2}$ mv² = Pt

$\Rightarrow$ v $\propto$ $\sqrt{t}$

$\frac{d x}{d t} = C \sqrt{t}$ [C = constant]

by integration.

$x = C \frac{t^{\frac{1}{2} + 1}}{\frac{1}{2} + 1}$

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

10. (JEE Main 2021 (Online) 18th March Morning Shift )

A constant power delivering machine has towed a box, which was initially at rest, along a horizontal straight line. The distance moved by the box in time 't' is proportional to :-

A. t^2/3

B. t^3/2

C. t

D. t^1/2

Correct Option is (B)

$P = F . v = m a v$

$P = \frac{m v d v}{d t}$

$\int_{0}^{t} P d t = m \int_{0}^{v} v d v$

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

$v = \sqrt{\frac{2 P t}{m}}$

$\frac{d x}{d t} = \sqrt{\frac{2 P t}{m}}$

$\int d x = \int \sqrt{\frac{2 P t}{m}} d t$

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