A light in extensible string that goes over a smooth fixed pulley as shown in the figure connect two blocks of masses `0.36kg and 0.72kg`. Taking `g = 10 m//s^(2)`. Find the work done by string on the block of mass `0.36 kg` during the first second after the system is released from rest.

Options

A. 4J
B. 2J
C. 8J
D. 10J

Answer

C

Explanation

So, acceleration = `("Net pulling foroce")/("Total mass")`
`=(0.72g-0.36g)/(0.72+0.36g)=(g)/(3)`
Distance , s = `ut + (1)/(2) at^(2)`
`= 0 + (1)/(2) xx (g)/(3)xx (1)^(2) = (g)/(6)`
So, T 0 mg = ma
T - 0.36 g = 0.36 a
T = 0.48 g
Now, work done by string on the block `W_(r) = T = cos^(@)` (on 0.36 kg of mass)
`=(0.48 g) ((g)/(6)) (1) =0.08 g^(2)`
`0.08 xx (10)^(2) = 8j`