Question

A block of mass 0.84 kg is suspended by a string which is wrapped so that it is at a radius of 0.061 m from the center of a pulley. The moment of inertia of the pulley is 6.20 times 10^-3 kg m^2. There is friction as the pulley turns. The block starts from rest, and its speed after it has traveled downwards a distance of D= 0.65 m, is 1.705 m/s. Calculate the amount of energy dissipated up to that point.

A) 5.620 times 10^-1
B) 8.150 times 10^-1
C) 1.182
D) 1.713
E) 2.484
F) 3.603
G) 5.224
H) 7.574

Answer 1

`E_l = 1.713 J`

Step-by-step explanation:

Given data:

mass of block is
`M_b = 0.84 kg`

radius of block = 0.061 m

moment of inertia is
`6.20 * 10^(-3) kg m^2`

D is distance covered by block = 0.65 m

speed of block is 1.705 m/s

From conservation of momentum we have

`M_b g D = (1)/(2) M_b v^2 + (1)/(2) I \omega^2 +  E_(loss)`

`0.84 * 9.81 * 0.65 = (1)/(2)*  0.84 * 1.705^2 +(1)/(2) * 6.2 * 10^(-3) [(1.705)/(0.061)]^2 + E_l`

solving for energy loss

`E_l = 1.713 J`