Question
A block of mass $$m$$ is connected to another block of mass $$M$$ by a spring (massless) of spring constant $$k.$$ The block are kept on a smooth horizontal plane. Initially the blocks are at rest and the spring is unstretched. Then a constant force $$F$$ starts acting on the block of mass $$M$$ to pull it. Find the force of the block of mass $$m.$$
A.
$$\frac{{MF}}{{\left( {m + M} \right)}}$$
B.
$$\frac{{mF}}{M}$$
C.
$$\frac{{\left( {M + m} \right)F}}{m}$$
D.
$$\frac{{mF}}{{\left( {m + M} \right)}}$$
Answer :
$$\frac{{mF}}{{\left( {m + M} \right)}}$$
Solution :
Drawing free body-diagrams for $$m\& M,$$
we get $$T = ma$$ and $$F - T = Ma$$
where $$T$$ is force due to spring
$$\eqalign{ & \Rightarrow F - ma = Ma\,{\text{or,}}\,F = Ma + ma \cr & \therefore a = \frac{F}{{M + m}} \cr} $$
Now, force acting on the block of mass $$m$$ is
$$ma = m\left( {\frac{F}{{M + m}}} \right) = \frac{{mF}}{{m + M}}.$$
Drawing free body-diagrams for $$m\& M,$$
we get $$T = ma$$ and $$F - T = Ma$$
where $$T$$ is force due to spring
$$\eqalign{ & \Rightarrow F - ma = Ma\,{\text{or,}}\,F = Ma + ma \cr & \therefore a = \frac{F}{{M + m}} \cr} $$
Now, force acting on the block of mass $$m$$ is
$$ma = m\left( {\frac{F}{{M + m}}} \right) = \frac{{mF}}{{m + M}}.$$
