Question

In $FIGURECP7.54$, find an expression for the acceleration of m1. The pulleys are massless and frictionless.

Hint: Think carefully about the acceleration constraint.

Short answer

The acceleration of block$m_1$is given by$a_1=\frac{2m_{2}g}{(2m_1+m_2)}$

Step-by-step solution

Given information

Mass of the block $1$is $m_1$

mass of the block $2$is $m_2$

The pulleys are massless and frictionless

The acceleration of the block $1$is localid="1649864857767" $a_1$

The acceleration of the block$2$is$a_2$

Explanation

Consider the Forces of two masses given in the$figureCP7.54$

For block of mass$m_1$

For block of mass$m_2$

$m_{2}g-T_2=m_2{a}_2$

Considering the pulley is massless,$\therefore T_1=T_2$

Let $a_2=a$$\therefore T=2m_{1}a$

$\because a_1=2a$acceleration constraint

On substituting the above equations on block equation, we get

$$\begin{gathered} m_{2}g-2m_{1}a=m_{2}a \\ (2m_1+m_2)a=m_{2}g \\ a=\frac{m_{2}g}{(2m_1+m_2)} \end{gathered}$$

Therefore, the acceleration for $m_1$is given by,

$a_1=\frac{2m_{2}g}{(2m_1+m_2)}$