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A block of mass M is pulled along a horizontal frictionless surface by a rope of mass m. If a force P is applied at the free end of the rope, the force exerted by the rope on the block is

Asked in JEE Main 2003 · Chains, ropes and internal forces

Answer: (4) (PM)/(M+m)

Step-by-step solution

Idea: the rope has mass, so part of P goes into accelerating the rope and only the rest reaches the block.

Treat rope and block as one body of mass M+m pulled by P:

a=P/(M+m).

Now isolate the block. The only horizontal force on it is the rope's pull T:

T=Ma=(PM)/(M+m).

Check the massless limit: as m→0, T→ P — the familiar result that a light rope passes the whole force along.

The heavier the rope, the more of P it keeps, which is exactly why real tow ropes are made light.

Why the other options are wrong

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