Question

If R = 12 cm, M = 310 g, and m = 40 g , find the speed of the block after it has descended 50 cm starting from rest. Solve the problem using energy conservation principles. (Treat the pulley as a uniform disk.)

Answer 1

`v = 1.42 m/s`

Step-by-step explanation:

While mass is falling downwards there is no frictional loss so here we can use mechanical energy conservation

So change in gravitational potential energy = gain in kinetic energy of the system

`mgh = (1)/(2)mv^2 + (1)/(2)I\omega^2`

for uniform cylinder we will have

`I = (1)/(2)MR^2`

now we have

`mgh = (1)/(2)I\omega^2 + (1)/(2)mv^2`

`mgh = (1)/(2)((1)/(2)MR^2)((v)/(R))^2 + (1)/(2)mv^2`

`mgh = (1)/(4)Mv^2 + (1)/(2)mv^2`

`mgh = ((M)/(4) + (m)/(2))v^2`

now we have

`v^2 = (mgh)/(((M)/(4) + (m)/(2)))`

`v^2 = (40(9.81)(0.50))/(((310)/(4) + (40)/(2)))`

`v = 1.42 m/s`