A light inextensible string that goes over a smooth fixed pulley as shown in the figure connects two blocks of masses 0.36kg and 0.72kg.Taking g=$10{ms}^{- 2}$ ,find the work done (in joule)by string on the block of mass 0.36kg during the first second after the system is released from rest . [image:6_image12.jpeg]

Question

Dr. Abubacker · Accepted Answer

(a) $a = \frac{Net\ pulling\ force}{Total\ mass}$
$= \frac{0.72g - 0.36g}{0.72 + 0.36} = \frac{g}{3}$
$s = \frac{1}{2}at^{2} = \frac{1}{2}\left( \frac{g}{3} ight)(1)^{2} = \frac{g}{6}$
T$- 0.36g = 0.36a = 0.36\frac{g}{3}$
$	herefore T = 0.48g$
Now,$w_{T} = TS\ \cos{0{^\circ}(on\ 0.36\ kg\ mass)}$
$= (0.48\ g)\left( \frac{g}{6} ight)(1) = 0.08(g^{2})$
$= 0.08(10)^{2} = 8J$
[image:6_image71.jpeg]

Q. A light inextensible string that goes over a smooth fixed pulley as shown in the figure connects two blocks of masses 0.36kg and 0.72kg.Taking g=$10{ms}^{- 2}$ ,find the work done (in joule)by string on the block of mass 0.36kg during the first second after the system is released from rest .

Explanation

Reviewed by Dr. Abubacker

Want more practice?