Answered

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 :

Answer:

[tex]E_l = 1.713 J[/tex]

Explanation:

Given data:

mass of block is [tex]M_b = 0.84 kg[/tex]

radius of block = 0.061 m

moment of inertia is [tex]6.20 \times 10^{-3} kg m^2[/tex]

D is distance covered by block = 0.65 m

speed of block is 1.705 m/s

From conservation of momentum  we have

[tex]M_b g D = \frac{1}{2} M_b v^2 + \frac{1}{2} I \omega^2 +  E_{loss}[/tex]

[tex]0.84 \times 9.81 \times 0.65 = \frac{1}{2}\times  0.84 \times 1.705^2 +\frac{1}{2} \times 6.2 \times 10^{-3} [\frac{1.705}{0.061}]^2 + E_l[/tex]

solving for energy loss

[tex]E_l = 1.713 J[/tex]

Other Questions