Physics Circle home Student Login

Home›Chapters›Laws of Motion PYQs›Friction Basics›Question 10

JEE Main 2023 (1 Feb, Shift 1) · Friction Basics

Friction Basics question from JEE Main 2023 (1 Feb, Shift 1)

Q10JEE Main 2023 · 1 FEB · SHIFT 1Medium

Friction Basics

A block of mass 5 kg5\text{ kg} is placed at rest on a table of rough surface. Now, if a force of 30 N30\text{ N} is applied in the direction parallel to surface of the table, the block slides through a distance of 50 m50\text{ m} in an interval of time 10 s10\text{ s}. Coefficient of kinetic friction is (given, g=10 ms−2g = 10\text{ ms}^{-2}):

Show answer and solution

Answer: (d)

  1. Starting from rest: s=12at2⇒50=12a(10)2⇒a=1 m/s2s = \tfrac{1}{2}at^2 \Rightarrow 50 = \tfrac{1}{2}a(10)^2 \Rightarrow a = 1\text{ m/s}^2
  2. F−μmg=ma⇒30−μ(50)=5(1)F - \mu mg = ma \Rightarrow 30 - \mu(50) = 5(1)
  3. μ=2550=0.5\mu = \dfrac{25}{50} = 0.5
  4. So the answer is (d) 0.50.
Report a mistake

More questions on Friction Basics