A body of mass 2 kg begins to move under the action of a time dependent force given by $\overrightarrow{\mathrm{F}}=\left(6 \mathrm{t} \hat{\mathrm{i}}+6 \mathrm{t}^{2} \hat{\mathrm{j}}\right) \mathrm{N}$. The power developed by the force at the time t is given by:

(D) $\left(9 t^{3}+6 t^{5}\right) W$

(A) $\left(6 t^{4}+9 t^{5}\right) W$

(B) $\left(3 t^{3}+6 t^{5}\right) W$

(C) $\left(9 t^{5}+6 t^{3}\right) W$

JEE Main · 2024 · 31 Jan 2024 · Shift 2 · Work, Power & Energy

A body of mass 2 kg begins to move under the action of a time dependent force given by

F = (6 t \hat{i} + 6 t^{2} \hat{j}) N

. The power developed by the force at the time t is given by:

(A) $(6 t^{4} + 9 t^{5}) W$
(B) $(3 t^{3} + 6 t^{5}) W$
(C) $(9 t^{5} + 6 t^{3}) W$
(D) $(9 t^{3} + 6 t^{5}) W$ ✓ Correct

Answer: (D)

Solution

From

F = m a

, the acceleration is

a = \frac{F}{m} = (3 t \hat{i} + 3 t^{2} \hat{j}) m/s^{2}

Starting from rest, velocity is obtained by integration:

v = \int_{0}^{t} a d t = \frac{3 t ^{2}}{2} \hat{i} + t^{3} \hat{j}

Power delivered by the force is

P = F \cdot v = (6 t) (\frac{3 t ^{2}}{2}) + (6 t^{2}) (t^{3}) = 9 t^{3} + 6 t^{5} W

JEE Main · 2024 · 31 Jan 2024 · Shift 2 · Single correct option

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