A small block with mass 0.0325 kg slides in a vertical circle of radius 0.600 m

Question

A small block with mass 0.0325 kg slides in a vertical circle of radius 0.600 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 4.00 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 0.660 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.

Wfr = ?

So far I think I solved for Vi = 8.44 m/s & Vf = 2.51 m/s

I keep getting work = -.864 J but that is wrong. Thanks`!

Explanation / Answer

normal reaction = weigth + centrifugal force

N= mg+mvi2/R

4= 0.0325*9.81 + 0.0325*v2/0.6

vi=8.24379m/s

for vf

at point B

N+mg = mvf2/R

vf =4.251 m/s

by energy coservation

K.Ei + P.Ei = K.E.f + P.Ef + friction loss(friction work done)

P.Ef =mgh where h =2*R because it is above height of 2R from intial state

0.5*0.0325*8.243792 + 0 = 0.5*0.0325*4.2512 +0.0325*9.81*1.2 + energy loss in friction

energy loss in friction = 0.42810J

above is energy loss due to friction so but work done by friction is negative so answere = -0.42810J

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