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Question

Two bodies m1and m2 are kept on a table with coefficient of friction 'μ' and are joined by a spring initially, the spring is in its relaxed state. Find out the minimum force F which will make the other block m2 move.(K is the spring  constant).


Solution

The correct option is C
Motion of m2 starts when,
kx=μ.m2.g
Where  x = elongation in the spring.
x=μ.m2.gk
The minimum force will be such that m1 has no kinetic energy. Applying work energy principle for m1.
x0(Fμm1gkx)dx=0F=[μm1g+12kx]=[μm1g+μm2g2]Fmin=μg.m1+m22μg

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