Laws of Motion

JEE Main level, single correct, medium

124 of 210

Question 124 of 210

A block of mass ' m\mathrm{m} ' is connected to another block of mass ' M\mathrm{M} ' by a spring (massless) of spring constant ' k\mathrm{k} '. The blocks are kept on a smooth horizontal plane. Initially the blocks are at rest and the spring is unstretched. Then a constant force 'F\mathrm{F}' starts acting on the block of mass 'M\mathrm{M}' to pull it. Find the force on the block of mass 'm\mathrm{m}'

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Answer: (C) mF(m+M)\frac{\mathrm{m F}}{\mathrm{(m+M)}}

Kx=ma=mFm+M\mathrm{Kx}=\mathrm{ma}=\frac{\mathrm{mF}}{\mathrm{m}+\mathrm{M}}

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