Question

If the system in the figure is released from rest in the configuration shown, find the velocity of the block $Q$ after it has fallen through a distance $10\text{meters}$. Given $P=Q=10\text{kg}$.

• A. $8\text{m/s}$
• B. $8.85\text{m/s}$
• C. $9.5\text{m/s}$
• D. $10\text{m/s}$

Answer 1

The correct option is B $8.85\text{m/s}$

When Q falls down $10\text{meters}$, the pulley A will fall down 5 metres and hence P will be lifted up by $5\text{metres}$.

Loss of potential energy of system $=(10\times 10-10\times 5)9.8=50\times 9.8\text{J}$
If $v$ be the velocity of $Q$ after falling $10$ metres, then $\frac{v}{2}$ must be the velocity of $P$ at that instant.
Kinetic energy gained by the system
$=\frac{1}{2}\times 10\times v^2+\frac{1}{2}\times 10\times {\left(\frac{v}{2}\right)}^2=\frac{25}{4}{v}^2$
Gain of kinetic energy = Loss of potential energy
$\therefore \frac{25}{4}{v}^2=50\times 9.8$
$\Rightarrow v=8.85\text{m/sec}$