Question

Two blocks of masses 400 g and 200 g are connected through a light string going over a pulley which is free to rotate about its axis. The pulley has a moment of inertia $1.6\times {10}^{-4}kg-m^2$ and a radius 2.0 cm. Find 'x', which is the kinetic energy of the system as the 400 g block falls through 50 cm and also find 'y', the speed of the blocks at this instant.

• A. x = 10.5 J, y = 2.6 m/s
• B. x = 9.8 J, y = 2.6 m/s
• C. x = 9.8 J, y = 1.4 m/s
• D. x = 0.98 J, y = 1.4 m/s

Answer 1

The correct option is D

x = 0.98 J, y = 1.4 m/s

According to the question

$0.4g-T_1=0.4a$ ...(1)

$T_2-0.2g=0.2a$ ....(2)

$(T_1-T_2)r=\frac{Ia}{r}$ ...(3)

From equation 1,2 and 3

$\Rightarrow a=\frac{(0.4-0.2)g}{\left(\frac{0.4+0.2+1.6}{0.4}\right)}=\frac{g}{5}$

Therefore (y) V=$\sqrt{2ah}=\sqrt{(2\times g{l}^5\times 0.5)}$

$\Rightarrow \sqrt{\left(\frac{g}{5}\right)}=\sqrt{\left(\frac{9.8}{5}\right)}=\frac{1.4m}{s}$.

(x) total kinectic energy of the system

$=\frac{1}{2}{m}_1{V}^2+\frac{1}{2}{m}_2{V}^2+\frac{1}{2}{18}^2$

$=(\frac{1}{2}\times 0.4\times {1.4}^2)+(\frac{1}{2}\times 0.2\times {1.4}^2)+(\frac{1}{2}\times (\frac{1.6}{4})\times {1.4}^2)=0.98joule$.