Physics Circle home Student Login

Home›Chapters›Laws of Motion PYQs›Two-Body Friction›Question 6

JEE Main 2021 (17 Mar, Shift 1) · Two-Body Friction

Two-Body Friction question from JEE Main 2021 (17 Mar, Shift 1)

Q6JEE Main 2021 · 17 MAR · SHIFT 1Medium

Two-Body Friction

Two blocks (m = 0.5 kg and M = 4.5 kg) are arranged on a horizontal frictionless table as shown in figure. The coefficient of static friction between the two blocks is 37\frac{3}{7}. Then the maximum horizontal force that can be applied on the larger block so that the blocks move together is ___________ N. (Round off to the Nearest Integer) [Take g as 9.8 ms−{}^{-}2]

Figure

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

Show answer and solution

Answer: 21

  1. m moves only through friction: amax=μg=37×9.8=4.2a_{max} = \mu g = \frac{3}{7} \times 9.8 = 4.2 m/s2^2.
  2. Fmax=(M+m)amax=5×4.2F_{max} = (M + m)a_{max} = 5 \times 4.2
  3. Fmax=21F_{max} = 21 N
Report a mistake

More questions on Two-Body Friction