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Physics Question 49 – JEE-MAIN 2026

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A body of mass 2 kg begins to move under the influence of time dependent force F→=(2ti^+6t2j^) N, where i^ and j^ are unit vectors along x and y-axis respectively. The power produced by the force at t=2 s is _______ W.

Power is the rate at which work is done, which can be expressed as the dot product of force and velocity.

● Video Walkthrough
Step 1: Determine acceleration and velocity as functions of time✦ Active

Given the force F→=(2ti^+6t2j^) N and mass m=2 kg. According to Newton's second law, acceleration is a→(t)=F→(t)m.

a→(t)=12(2ti^+6t2j^)=(ti^+3t2j^) m/s2

Since the body "begins to move", its initial velocity v→(0)=0→. Velocity is the integral of acceleration with respect to time.

v→(t)=∫a→(t)dt=∫(ti^+3t2j^)dt=(t22i^+t3j^) m/s
Step 2: Calculate force and velocity at t=2 s○ Expand

Substitute t=2 s into the expressions for force and velocity.

F→(2)=(2(2)i^+6(2)2j^)=(4i^+6(4)j^)=(4i^+24j^) N
v→(2)=((2)22i^+(2)3j^)=(42i^+8j^)=(2i^+8j^) m/s
Step 3: Compute the instantaneous power at t=2 s○ Expand

The instantaneous power P is the dot product of the force and velocity vectors.

P(2)=F→(2)⋅v→(2)=(4i^+24j^)⋅(2i^+8j^)
P(2)=(4)(2)+(24)(8)=8+192=200 W
💡 Teacher's Secret Hint

Ensure to perform the dot product correctly by multiplying corresponding components and summing them.

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