Question

Two masses m and m' are tied with a thread passing over a pulley, m' is on a frictionless horizontal surface and m is hanging freely. If the acceleration due to gravity is g, the acceleration of m' in this arrangement will be (A)g (B)mg/(m+m') (C)m'g/(m+m') (D) g(m-m')/(m+m')

Answer 1

Net pulling force= mg.
Total mass being pulled=(m+m').
hence acceleration (a) of both the block is given by:
a= (net pulling force)/(total mass)
=(mg)/(m+m').

Answer 2

Answer:

The acceleration of a =$\frac{mg}{(m+m')}$

Explanation:

Frictional Force:

A force that prevents one solid object from slipping or rolling over another. Frictional forces, such as the traction required to walk without slipping, can be advantageous, but they also create significant resistance to motion.

Net pulling power=mg

The total mass being dragged=(m+m')

For mass m , T=ma------------------(1)

For mass m' , m'g-T=m'a

                      m'g=(m+m')a

                         a=$\frac{m'g}{(m+m')}$

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