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JEE Main 2007 (AIEEE) · Connected Bodies & Pulleys

Connected Bodies & Pulleys question from JEE Main 2007 (AIEEE)

Q41JEE Main 2007 · AIEEEMedium

Connected Bodies & Pulleys

A block of mass mm is connected to another block of mass MM by a spring (massless) of spring constant k.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 FF starts acting on the block of mass MM to pull it. Find the force on the block of mass m.m.

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Answer: (d)

  1. Treat both blocks as moving with a common acceleration a=Fm+Ma = \dfrac{F}{m+M}.
  2. The only horizontal force on mm is the spring force, so Fm=maF_m = ma.
  3. Fm=mFm+MF_m = \dfrac{mF}{m+M}
  4. So the answer is (d) mF(m+M)\dfrac{mF}{(m+M)}.
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