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Physics Question 49 – JEE-MAIN 2026

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.

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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