JEE Main Work, Energy and Power PYQ

26. (JEE Main 2023 (Online) 24th January Morning Shift )

A spherical body of mass 2 kg starting from rest acquires a kinetic energy of 10000 J at the end of $5^{th}$ second. The force acted on the body is ________ N.

Correct answer is (40)

Let the force be $F$ so acceleration $a = \frac{F}{m}$

So displacement $S = \frac{1}{2} a t^{2} = \frac{F t^{2}}{2 m}$

So work done $W = F . S = \frac{F^{2} t^{2}}{2 m}$

From work energy Theorem

$Δ K E = W$

$W = \frac{F^{2} t^{2}}{2 m} = 10000$

$F = \sqrt{\frac{10000 \times 2 \times 2}{5^{2}}}$

$F = 40 N$

27. (JEE Main 2022 (Online) 28th July Evening Shift)

A bullet of mass $200 g$ having initial kinetic energy $90 J$ is shot inside a long swimming pool as shown in the figure. If it's kinetic energy reduces to $40 J$ within $1 s$ , the minimum length of the pool, the bullet has to travel so that it completely comes to rest is

JEE Main 2022 (Online) 28th July Evening Shift Physics - Work Power & Energy Question 27 English

A. 45 m

B. 90 m

C. 125 m

D. 25 m

Correct Option is (A)

$\frac{1}{2} m x^{2} = 90$

$\Rightarrow \frac{1}{2} \times 0.2 \times x^{2} = 90$ ,

$x^{2} = 900$

$x = 30$ m/s

$\frac{1}{2} m v^{2} = 40 \Rightarrow v = \frac{2}{3} \times 30 = 20$ m/s

$20 = 30 - a \times 1 \Rightarrow a = - 10$ m/s²

$0 - x^{2} = 2 a s$

$s = \frac{x^{2}}{- 2 a} = \frac{30 \times 30}{2 \times 10}$

$= 45$ m

28. (JEE Main 2022 (Online) 26th July Morning Shift )

As per the given figure, two blocks each of mass $250 g$ are connected to a spring of spring constant $2 {Nm}^{- 1}$ . If both are given velocity $v$ in opposite directions, then maximum elongation of the spring is :

JEE Main 2022 (Online) 26th July Morning Shift Physics - Work Power & Energy Question 30 English

A. $\frac{v}{2 \sqrt{2}}$

B. $\frac{v}{2}$

C. $\frac{v}{4}$

D. $\frac{v}{\sqrt{2}}$

Correct Option is (B)

$∵$ Loss in KE = Gain in spring energy

$\Rightarrow \frac{1}{2} m v^{2} \times 2 = \frac{1}{2} k x_{m}^{2}$

$\Rightarrow 2 \times \frac{1}{4} \times v^{2} = 2 \times x_{m}^{2}$

$\Rightarrow x_{m} = \sqrt{\frac{v^{2}}{4}} = \frac{v}{2}$

29. (JEE Main 2022 (Online) 25th July Evening Shift )

A bag of sand of mass 9.8 kg is suspended by a rope. A bullet of 200 g travelling with speed 10 ms^{$-$ 1} gets embedded in it, then loss of kinetic energy will be :

A. 4.9 J

B. 9.8 J

C. 14.7 J

D. 19.6 J

Correct Option is (B)

Loss in $K E = \frac{1}{2} \times \frac{m_{1} m_{2}}{m_{1} + m_{2}} \times v^{2}$

$= \frac{1}{2} \times \frac{9.8 \times 0.2}{10} \times (10)^{2}$

$= 9.8$ J

30. (JEE Main 2022 (Online) 25th July Morning Shift )

A body of mass $0.5 kg$ travels on straight line path with velocity $v = (3 x^{2} + 4) m / s$ . The net workdone by the force during its displacement from $x = 0$ to $x = 2 m$ is :

A. 64 J

B. 60 J

C. 120 J

D. 128 J

Correct Option is (B)

$v = 3 x^{2} + 4$

at $x = 0$ , $v_{1} = 4$ m/s

$x = 2$ , $v_{2} = 16$ m/s

$\Rightarrow$ Work done = $Δ$ kinetic energy

$= \frac{1}{2} \times m (v_{2}^{2} - v_{1}^{2})$

$= \frac{1}{4} (256 - 16)$

$= 60$ J