Question

A block of mass M is pulled along a horizontal frictionless surface by a rope of mass m by applying a force P at one end of the rope. The force which the rope exerts on the block is...?

Answer is PM/(M+m)

Answer 1

We know that F=kma (Newton's second law of motion)

Now, (M+m)a=P -(1) as they are connected

Let the force exerted on M be x.

Then Ma=x - (2)

Now (1)/(2) will give you x which is PM/M+m.

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Answer 2

Answer: PM/(M + m)

Explanation:

(ΣFx ≠ 0)

Let the net acceleration of the whole system be = a

By the FBDs,

1. P - T = ma
2. T = Ma

Adding both,

P - T + T = Ma + ma

=> P = (M + m)a

=> a = P/(M + m) _3

Now, we are asked to find the tension.

∴ T = M.a = PM/(M + m) [From 3].

More:

• Tension in a string is same in every point (massless).
• Tension (for rope having mass) is maximum at the point which is under greatest influence by an external force. In the other end , T = 0.
• T = F/L(L - x)

where, F = force, L = original length of string, x = dist from the end.

• Unlike springs, strings or ropes couldn't change their shape.