A block of mass $m_{1}$ is kept over another block of mass $m_{2}$ and the system rests on a horizontal surface (as shown in figure) . A constant horizontal force 2F acting on the block produces an acceleration $\frac{F}{m_{1} + m_{2}}$ in the system, the two blocks always move together .(a) Find the coefficient of kinetic friction between the bigger block and the horizontal surface .(b) Find the frictional force acting on the smaller block . (c) Find the work done by the force of friction on the smaller block by the bigger block during a displacement x of the system.

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Solution

a) Frictional force,

$f = (m_{1} + m g) μ g$

As shown in figure;

$2 F - f = (m_{1} + m_{2}) a$

$2 F - f = (m_{1} + m_{2}) . \frac{F}{(m_{1} + m_{2})}$

$F = f$

b) $F_{1} = m_{1} a$

$F_{1} = \frac{m_{1} F}{m_{1} + m_{2}}$

c) Since both block move together hence no any displacement between $m_{1}$ and $m_{2}$ and hence work done is zero.

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a) Frictional force,f=(m1+mg)μgAs shown in figure;2F−f=(m1+m2)a2F−f=(m1+m2).F(m1+m2)F=fb) F1=m1aF1=m1Fm1+m2c) Since both block move together hence no any displacement between m1 and m2 and hence work done is zero.