Question

A small block with mass 0.0475 kg slides in a vertical circle of radius 0.550 m on the inside of a circular track. During one of the revolutions of the block, when the block is at the bottom of its path, point A, the magnitude of the normal force exerted on the block by the track has magnitude 3.85 N . In this same revolution, when the block reaches the top of its path, point B, the magnitude of the normal force exerted on the block has magnitude 0.695 N .

How much work was done on the block by friction during the motion of the block from point A to point B?

Express your answer with the appropriate units.

Answer 1

at the bottom

Nbot - m*g = m*vbot^2/r

3.85 - (0.0475*9.8) = 0.0475*vbot^2/0.55

vbot^2 = 6.26 m/s

at the top

Ntop + m*g = m*vbot^2/r

0.695 + (0.0475*9.8) = 0.0475*vtop^2/0.55

vtop^2 = 3.66 m/s

from work energy relation

total work done = change in KE

Wg + Wf = dKE

Wg = UA - UB

uA = 0

UB = m*g*h = m*g*2r

-m*g*2r + Wf = (1/2)*m*(vtop^2 - vbot^2)

-0.0475*9.8*2*0.55 + Wf = (1/2)*0.0475*(3.66^2-6.26^2)

work done by friction Wf = -0.1 J

DONE please check the answer. any doubts post in comment box