Question

The pulley shown in the figure has a radius of 20 cm and moment of inertia 0.2 kg-$m^2$. The string going over it is attached at one end to a vertical spring of spring constant 50 N/m fixed from below, and supports a 1 kg mass at the other end. The system is released from rest with the spring at its natural length. Find the speed of the block when it has descended through 10 cm. Take g = 10 $m/s^2$

.

• A. 1 m/s
• B. 0.5m/s
• C. 2m/s
• D. zero

Answer 1

The correct option is B

0.5m/s

I = 0.2 kg-$m^2$, r = 0.2 m, K=50 N/m,

m=1 kg, g=10 $m{s}^{-2}$, h=0.1 m

Therefore applying the law of conservation of energy

mgh = $\frac{1}{2}m{v}^2+\frac{1}{2}k{x}^2+\frac{1}{2}I\left(\frac{v^2}{R^2}\right)$

$\Rightarrow 1=\frac{1}{2}\times 1\times V^2+\frac{1}{2}\times \frac{0.2\times \frac{V^2}{0.04}}{2}+\left(\frac{1}{2}\right)\times 50\times 0.01(x=h)$

$\Rightarrow 1=0.5v^2+2.5v^2+\frac{1}{4}$

$\Rightarrow 3v^2=\frac{3}{4}$

$\Rightarrow v=\frac{1}{2}=0.5m/s$.