Question

Find the acceleration of the two masses as shown in figure. The pulleys are light and frictionless and strings are light and inextensible.

Answer 1

Given, strings are light and inextensible, therefore, tension remains same throughout the system be $T$.

Let,

Acceleration of block of mass m be $a_1$

Acceleration of block of mass M be $a_2$

Consider block of mass M,

from F.B.D.,

$5T-Mg=M{a}_2$

$5T=M(g+a_2)$ eqn 1

Now from F.B.D of block of mass m,

$mg-T=m{a}_1$

$T=m(g-a_1)$ eqn 2

From constraint equation,

$(\sum )T.a=0$

therefore,

$\left(5T\right).\left(a_2\right)+\left(T\right).(-a_1)=0$ ( since tension and acceleration are in opposite direction)

$5a_2=a_1$ eqn 3

On solving 1 and 2,

$5\left(m\right(g-a_1\left)\right)=M(g+a_2)$

also, $a_1=5a_2$

Therefore,

$5\left(m\right(g-5a_2\left)\right)=M(g+a_2)$

on solving , we'll get

$5mg-25m{a}_2=Mg+M{a}_2$

$5mg-Mg=(25m+M)a_2$

therefore,

ANS:

$a_2=\frac{5mg-Mg}{(25m+M)}$

also,

$a_1=5\frac{5mg-Mg}{(25m+M)}$