Question

Define impulse. Two blocks of masses $10g$ and $30g$ are fastened to the ends of a cord which passes over a frictionless pulley. The $10g$ block rests on the table. What minimum force must be applied on the $10g$ block to keep it on the table? What will be the tension in the cord under this condition? What will be the acceleration of the system and the tension in the cord when the force is withdrawn?
($g=9.8m{s}^{-2}$)

Answer 1

consider the weight $30gm$ is going to downward with the acceleration be $a$.

And the tension acting in upward direction.

Now,

$f=m_{1}g-T$

$m_{1}a=m_{1}g-T$

and $m_1=30gm$

And the $10gm$ object will also move with the same acceleration.

Then, $T=m_{2}a$ and $m_2=10gm$

thus, $m_{1}a=m_{1}g-m_{2}a$

Therefore,

$\frac{m_{2}g}{m_1+m_2}$

$a=\frac{30\times 10}{10+30}$

$a=\frac{300}{40}=\frac{30}{4}=7.5$

Hence,

$a=7.5m/s^2$

and the tension $T$ in the string $=\frac{10}{1000}7.5$

$T=0.075N$

Hence, Force required to stop the object of $10gm$ mass is $0.075N$