Question

What is the acceleration of the $3.0kg$block in $FIGURECP7.55$across the frictionless table?

Hint: Think carefully about the acceleration constraint.

Short answer

The acceleration of the $3.0kg$block is $2.8m/s^2$

Step-by-step solution

Given information

mass of block$m_1=3.0kg$

mass of block$m_2=1.0kg$

All surfaces are frictionless.

Explanation

Consider the Forces of two masses given in the$figureCP7.55$

For block mass$m_1$

$T_1=m_1{a}_1$

For block 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 mass$m_2$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,

localid="1649865147826" $a_1=\frac{2m_{2}g}{(2m_1+m_2)}=\frac{2(1.0)(9.81)}{\left(2\right(3.0)+1.0)}=2.8m/s^2$