Question

The pulley shown in figure has radius $20cm$ and moment of Inertia $0.2kg{m}^2$. Initially the spring with a constant $50N/m$ is relaxed. The system is released from rest. Then the velocity of $1kg$ block when it has descended by $10cm$ is :

(Assume there is no sliding between pulley and string and $g=10m/s^2$)

• A. $1m/s$
• B. $\frac{1}{2}m/s$
• C. $2m/s$
• D. None of above

Answer 1

The correct option is B $\frac{1}{2}m/s$

Applying conservation of energy,

Change of potential energy of block = spring energy + Rotational energy of pulley + kinetic energy of the block.

$Mg(x-0)=\frac{1}{2}k{x}^2+\frac{1}{2}I{\omega }^2+\frac{1}{2}M{v}^2$

$1\times 10\times 0.1=\frac{1}{2}\times 50(0.1{)}^2+\frac{1}{2}(0.2){\left(\frac{V}{0.2}\right)}^2+\frac{V^2}{2}]$ ,as$(k=50N/m,x=0.1m,\omega =\frac{V}{0.2})$

$1=\frac{1}{4}+3V^2$

$3V^2=\frac{3}{4}$

$\Rightarrow V=\frac{1}{2}m/s$