Question

pls answer this question.....A block of mass M placed on a frictionless
horizontal table is pulled by another block of
mass m hanging vertically by a massless string
passing over a frictionless pulley. The tension
in the string is?​

Answer 1

Answer:

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Answer 2

The tension in the string is  $\frac{Mmg}{M+m}$.

Explanation:

Please refer to the diagram in the attachment.

The block of mass 'M' is moving to the right and block of mass 'm' is moving down.

Let us say the magnitude of acceleration of both the blocks is 'a'.

The normal force(N) on block of mass 'M' balances gravitational force on it.(∴ there is no acceleration to it in vertical direction)

∴ N = Mg

The only horizontal force on block of mass 'M' is Tension(T).

∴ T = Ma

On block of mass 'm' the 2 vertical forces acting on it are Tension(upwards) and gravitational force(downwards).

Net acceleration of these 2 forces is 'a' in downward direction.

∴ mg - T = ma

Substitute T = M×a in this equation.

mg -  Ma = ma

mg = (M+m)a

∴ a = $\frac{mg}{M+m}$

Now substitute this acceleration(a) in T = M×a.

∴ T  = $\frac{Mmg}{M+m}$