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Laws of Motion question

2007 · Shift 0 · Q97
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Laws of Motion question

2007 · Shift 0 · Q97

JEE MainPhysicsLaws of MotionMCQ+4 / −1
A block of mass mmm is connected to another block of massMmassMmassM by a spring (massless) of spring constant k.k.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 FFF starts acting on the block of mass MMM to pull it. Find the force of the block of mass m.m.m.
  1. A
    MF(m+M){{MF} \over {\left( {m + M} \right)}}(m+M)MF​
  2. B
    mFM{{mF} \over M}MmF​
  3. C
    (M+m)Fm{{\left( {M + m} \right)F} \over m}m(M+m)F​
  4. D
    mF(m+M){{mF} \over {\left( {m + M} \right)}}(m+M)mF​
View written solutionFree

Correct answer: D

  1. Understand what is being asked

A force FFF is applied on block MMM. The two blocks mmm and MMM are connected by a massless spring and move on a smooth horizontal surface.

We need the force on block of mass mmm. Since the only horizontal force acting on block mmm is the spring force, this means we need the spring force on mmm.

  1. Find the acceleration of the system

Treat both blocks together as one system.

  • Total mass =m+M= m+M=m+M
  • External horizontal force on the system =F= F=F

So, using Newton's second law,

a=Fm+Ma = \frac{F}{m+M}a=m+MF​
  1. Analyze block mmm separately

For block mmm, the only horizontal force is the spring force TTT.

Since block mmm moves with acceleration aaa,

T=maT = maT=ma

Substitute a=Fm+Ma = \dfrac{F}{m+M}a=m+MF​:

T=m(Fm+M)T = m\left(\frac{F}{m+M}\right)T=m(m+MF​)

Therefore,

T=mFm+MT = \frac{mF}{m+M}T=m+MmF​

This is the force acting on block mmm.

  1. Check options
  • A: MFm+M\dfrac{MF}{m+M}m+MMF​ ❌
  • B: mFM\dfrac{mF}{M}MmF​ ❌
  • C: (M+m)Fm\dfrac{(M+m)F}{m}m(M+m)F​ ❌
  • D: mFm+M\dfrac{mF}{m+M}m+MmF​ ✅
  1. Final answer

The force on block mmm is

mFm+M\boxed{\frac{mF}{m+M}}m+MmF​​

So the correct option is D.

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