First slide

Two block problem

Question

Block A of mass m rests on the plank B of mass 3m which is free to slide on a frictionless horizontal surface. The coefficient of friction between the block and plank is 0.2. If a horizontal force of magnitude 2 mg is applied to the plank B, the acceleration of A relative to the plank and relative to the ground respectively, are

Moderate

A

0, $\frac{g}{2}$
B

0, $\frac{2 g}{3}$
C

$\frac{3 g}{5}, \frac{g}{5}$
D

$\frac{2 g}{5}, \frac{g}{5}$

Solution

Block A moves due to friction. Maximum value of friction can be $μ m_{A} g$ . Therefore, maximum acceleration of A can be $\frac{μ m_{A} g}{m_{A}} or μ g = 0.2 g = \frac{g}{5}$ . When force 2 mg is applied on lower block, common acceleration (if both move together) will be
$a = \frac{Net force}{Total mass} = \frac{2 m g}{4 m} = \frac{g}{2}$
Since, a = g/2 is greater than maximum acceleration of A which can be given to it by friction. Therefore, slipping will take place.
$a_{A} = 0.2 g = \frac{g}{5}$
$a_{B} = \frac{2 m g - 0.2 m g}{3 m} = 0.6 g = \frac{3 g}{5}$
$a_{A B} = a_{A} - a_{B} = (\frac{g}{5} - \frac{3 g}{5})$
$= \frac{- 2 g}{5} (in backward direction)$

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