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 horizo-ntal 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

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

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

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

Solution

Maximum force applied on B for which A and B will move together is given by ${F_{\max }} = \mu .{m}g\left( { 1+ \frac{{{3m}}}{{{m}}}} \right)$ ${F_{\max }} = \mu .{m}g\left( { 1+ \frac{{{3m}}}{{{m}}}} \right)=0.2Xmg(1+3)=\frac {4}{5}mg$
Since applied force (2mg) is greater than F_max, block A will slip over B.
$\therefore a_A =\frac {\mu mg}{m}=\frac {g}{5}$ , $\therefore a_B =\frac {2mg-\mu mg}{3m}=\frac {3g}{5}$
$\therefore a_{A/B} =a_A-a_B=-\frac {2g}{5}$

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