A block with mass m = 10 kg slides down from rest along a surface that is inclin

Question

A block with mass m = 10 kg slides down from rest along a surface that is inclined 30° to the horizontal. There is kinetic friction between the block and the surface. The coefficient of kinetic friction is 0.10 . A string that is attached to the block is wrapped around a disk of mass M = 20 kg and radius of 0.2 m. The string doesn’t slip against the disk.

(a) If 10 cm of the string is unwound, find the velocity of the block. (ANSWER: 0.643 m/s)

(b) If 10 cm of the string is unwound, find the tension in the string. (ANSWER: 20.7 N)

Explanation / Answer

Given:

m = 10kg
uk = 0.10
M = 20 kg
r = 0.2 m
x = 10.0 cm = 0.1 m

Using Energy conservation,
Initial Potential Energy = Final kinetic Energy(Rotational + Translational) + Work done against friction
m*g*h = 1/2 * I * (v/r)^2 + 1/2*m*v^2 + uk*m*g*cos(30)*x
10*9.8*0.1*sin(30) = 1/2 * 20 * v^2 + 1/2 * 10 * v^2 + 0.1*10*9.8*cos(30)*0.1

Calculate
v = 0.519 m/s
Velocity of the block , v = 0.519 m/s Final answer

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