Physics Circle home Student Login

Home›Chapters›Laws of Motion PYQs›Inertia, First & Second Laws›Question 24

JEE Main 2021 (26 Feb, Shift 1) · Inertia, First & Second Laws

Inertia, First & Second Laws question from JEE Main 2021 (26 Feb, Shift 1)

Q24JEE Main 2021 · 26 FEB · SHIFT 1Medium

Inertia, First & Second Laws

A boy pushes a box of mass 2 kg with a force F⃗=(20i^+10j^) N\vec{F} = \left(20\hat{i} + 10\hat{j}\right)\text{ N} on a frictionless surface. If the box was initially at rest, then ___________ m is displacement along the x-axis after 10s.

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

Show answer and solution

Answer: 500

  1. ax=Fxm=202=10 m/s2a_x = \frac{F_x}{m} = \frac{20}{2} = 10\text{ m/s}^2
  2. x=12axt2=12(10)(10)2=500 mx = \frac{1}{2}a_xt^2 = \frac{1}{2}(10)(10)^2 = 500\text{ m}
  3. So the displacement along xx is 500500 m.
Report a mistake

More questions on Inertia, First & Second Laws