JEE Main 2025 Jan 28 Shift 2, Physics Q14: Work Energy Theorem And Its Applications

JEE Main2025Jan 28, Shift 2Physics

A body of mass 4 kg is placed on a plane at a point $P$ having coordinate $(3, 4)$ m. Under the action of force $F = (2 \hat{i} + 3 \hat{j})$ N, it moves to a new point $Q$ having coordinates $(7, 10)$ m in 4 sec. The average power and instantaneous power at the end of 4 sec are in the ratio of :

A
$13 : 6$
B
$6 : 13$
C
$1 : 2$
D
$4 : 3$

Solution

Displacement: $s = (7 \hat{i} + 10 \hat{j}) - (3 \hat{i} + 4 \hat{j}) = 4 \hat{i} + 6 \hat{j}$ m.

Work done: $W = F \cdot s = (2 \hat{i} + 3 \hat{j}) \cdot (4 \hat{i} + 6 \hat{j}) = 8 + 18 = 26$ J.

Average power: $⟨ P ⟩ = \frac{W}{t} = \frac{26}{4} = 6.5$ W.

Acceleration: $a = \frac{F}{m} = \frac{1}{2} \hat{i} + \frac{3}{4} \hat{j}$ m/s $^{2}$ .

Since the body starts from rest, velocity at $t = 4$ s: $v = a t = 2 \hat{i} + 3 \hat{j}$ m/s.

Instantaneous power at $t = 4$ s: $P_{inst} = F \cdot v = (2 \hat{i} + 3 \hat{j}) \cdot (2 \hat{i} + 3 \hat{j}) = 4 + 9 = 13$ W.

$∴ \frac{⟨ P ⟩}{P _{inst}} = \frac{6.5}{13} = \frac{1}{2}$ , so the ratio is $1 : 2$ .

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JEE Main 2025 Jan 28 Shift 2, Physics Q14: Work Energy Theorem And Its Applications

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