First slide

Two block problem

Question

A block B of mass 5kg is placed on a slab A of mass 20kg which lies on a frictionless surface as shown in the figure. The coefficient of static friction between the block and the slab is 0.4 and that of kinetic friction is 0.2. If a force F = 25N acts on B, the acceleration of the slab will be

Moderate

A

0.4ms^–2
B

0.5ms^–2
C

1ms^–2
D

zero

Solution

Maximum force upto which A and B will move together is given by. F_max = µ_smg(1+m/M), where m = mass of B and M = mass of A.
$\therefore$ F_max = 0.4 x 5 x 10 (1+ 5/20)N ⇒ 25N so A and B will move together and their acceleration is $a = \frac{{25}}{{20 + 5}}m/{s^2} = 1m/{s^2}$

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