A body of mass 1kg begins to move under the action of a time dependent force F⃗ = ( tî + 3t² ĵ ) N, where î and ĵ are the unit vectors along x and y axis. The power developed by above force, at the time t = 2s, will be ____________ W.

Solution

Physics sir · 4 steps

Power is F⃗ · v⃗, so first integrate the force to get the velocity at t = 2 s.

1. Acceleration. With m = 1 kg, a⃗ = F⃗ / m = t î + 3t² ĵ. The acceleration changes with time, so v = at does not apply.
2. Integrate for velocity. The body starts from rest, so v⃗ = ∫₀ᵗ a⃗ dt = t²/2 î + t³ ĵ. At t = 2 s this gives v⃗ = 2 î + 8 ĵ m s⁻¹.
3. Force at that instant. At t = 2 s, F⃗ = 2 î + 3(2)² ĵ = 2 î + 12 ĵ N.

   P = F⃗ · v⃗ = (2)(2) + (12)(8) = 4 + 96 = 100 W

4. So. The power developed at t = 2 s is 100 W.

The common slip: 296. Treating the acceleration as constant and using v⃗ = a⃗ t = 4 î + 24 ĵ gives 8 + 288 = 296, not 100.

The idea to keep. Instantaneous power is F⃗ · v⃗ at that instant. When the force depends on time, integrate the acceleration to get v⃗. Never use v = at for a changing acceleration.

Answer from the NTA final answer key. Solution by PrepWiser.