JEE Main 2023 · 31 January · evening · Single correct · +4 / −1

A body of mass 10 kg is moving with an initial speed of 20 m / s. The body stops after 5 s due to friction between body and the floor. The value of the coefficient of friction is:
(Take acceleration due to gravity g=10 ms⁻² )

  1. (a) 0.3
  2. (b) 0.2
  3. (c) 0.5
  4. (d) 0.4
Answer and solution

Answer (d) 0.4

Solution

Physics sir · 3 steps

Get the deceleration from the stopping time, then friction is the only force supplying it, so a = μg.

  1. Find the deceleration. The speed falls from 20 m s⁻¹ to 0 in 5 s, so the magnitude of the deceleration is a = 20/5 = 4 m s⁻².
  2. Friction does it all. On a level floor the normal force is N = mg. The only horizontal force is kinetic friction μ mg, so Newton's second law gives μ mg = ma and the mass cancels.

    μ = a/g = 4/10 = 0.4

  3. So. The coefficient of friction is 0.4, option (d).

The idea to keep. On a level surface with friction alone slowing a body, the deceleration is μ g. Mass never matters, so find a from kinematics and divide by g.

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

Study Friction

Asked about 40 times in JEE Main since 2020. This question is from 2023.