Physics Circle home Student Login

Home›Chapters›Work, Energy and Power PYQs›Power›Question 19

JEE Main 2023 (24 Jan, Shift 2) · Power

Power question from JEE Main 2023 (24 Jan, Shift 2)

Q19JEE Main 2023 · 24 JAN · SHIFT 2Medium

Power

A body of mass 1kg begins to move under the action of a time dependent force F⃗=(ti^+3t2j^)\vec{F} = \left(t\hat{i} + 3t^2\hat{j}\right) N, where i^\hat{i} and j^\hat{j} are the unit vectors along xx and yy axis. The power developed by above force, at the time t = 2s, will be ____________ W.

Numerical answer type. Work it out, then open the answer.

Show answer and solution

Answer: 100

  1. m=1m = 1 kg, so a⃗=ti^+3t2j^\vec{a} = t\hat{i} + 3t^2\hat{j}; from rest v⃗=t22i^+t3j^\vec{v} = \dfrac{t^2}{2}\hat{i} + t^3\hat{j}.
  2. At t=2t = 2: F⃗=2i^+12j^\vec{F} = 2\hat{i} + 12\hat{j}, v⃗=2i^+8j^\vec{v} = 2\hat{i} + 8\hat{j}.
  3. P=F⃗⋅v⃗=4+96=100 WP = \vec{F} \cdot \vec{v} = 4 + 96 = 100\text{ W}
Report a mistake

More questions on Power