Question

If R = 12 cm, M = 520 g, and m = 20 g (below), 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 = 0.84 m/s

Step-by-step explanation:

given,

R = 12 cm

M (mass of pulley )= 520 g

m (mass of block)= 20 g

s = 50 cm = 0.5 m

using conservation of energy

Potential energy = Kinetic energy

`m g h = (1)/(2)mv^2 + (1)/(2)I\omega^2`

`I_(disk)= (1)/(2)MR^2` and v = r ω

`m g h = (1)/(2)mv^2 + (1)/(2)((1)/(2)MR^2)((v)/(R))^2`

`m g h = (1)/(2)mv^2 +(1)/(4)Mv^2`

`m g h = (1)/(2)v^2(m +(1)/(2)M)`

`v=\sqrt{(2mgh)/(m + 0.5 M)}`

`v=\sqrt{(2* 0.020 * 9.8 * 0.5)/(0.02 + 0.5* 0.52)}`

v = √0.7

v = 0.84 m/s