Question

In figure $m_1=5$ kg and F = 1N. Find the acceleration of either block. Describe the motion of $m_1$ if the string breaks but F continues to act.

Answer 1

From the free body diagram (Figure)

$T+m_{1}a-(m_{1}g+F)=0$

$\Rightarrow T=m_{1}g+F-m_{1}a$

From the free body diagram (Figure)

$T+(m_{2}g+F+m_{2}a)=0$

$\Rightarrow T=m_{2}g+F+m_{2}a$

$\Rightarrow T=5g+1-5a$ ....(i)

$\Rightarrow T=2g+1+2a$ ....(ii)

From equation (i) and (ii) , we have

5g + 1- 5a = 2g + 1 + 2a

$\Rightarrow 3g-7a=0$

$\Rightarrow 7a=3g$

$\Rightarrow a=\frac{3g}{7}=\frac{29.4}{7}$

= 4.2 $m/s^2$ [g = $9.8m/s^2$]

Hence acceleration of block is $4.2m/s^2$.

After the string breaks, $m_1$ moves downward with force F acting downward, then ,

$m_{1}a=f+m_{1}g$

where force = 1 N and

acceleration = $\frac{1}{5}=0.2m/s^2$

$\therefore m_{1}a=(1+5g)$

So, acceleration $=\frac{\text{Force}}{\text{Mass}}$

$=\frac{5(g+0.2)}{5}$

=(g+ 0.2) $m/s^2$