Question

Figure shows a system of two blocks of 10 kg and 5 kg mass, connected by ideal strings and pulleys. Here ground is smooth and friction coefficient between the two blocks is µ = 0.5. A horizontal force F is applied on lower block as shown. The minimum value of F required to start sliding between the blocks is : (Tage g = 10 m/s2 )

(A) 12.5 N (B) 25 N (C) 50 N (D) 100 N

Answer 1

Correct option (B) 25 N

Explanation

Here at the instant of sliding limiting friction f = µN will act on the two blocks for limiting equilibrium,

we use :

F = T + µ N … (1)

and   T = µ N … (2)

and   2T + N = 50 … (3)

on solving (2) and (3), we get

N = 50/2µ + 1 = 25N

Now from (1) & (2)

F = 2 µN

= 2 × 0.5 × 25 = 25N.