A small block with mass 0.0475 kg slides in a vertical circle of radius 0.600 m

Question

A small block with mass 0.0475 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 3.80 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 .

Part A

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.

Explanation / Answer

Take height,h at point A as 0 and the height, h at point B will be,

r=d/2

d=2r then

d=2*0.6 m=1.2 m
now we need the v's at point A and point B
at point A

the normal force is,
F = m v^2 / r + m g
3.8 N = (0.0475 kg* v^2 / 0.600 m) + (0.0475 kg)(9.81 m/s2)
3.8 N = 0.0791 v^2 + 0.4659
0.0791 v^2 = 3.3341
v^2 = 42.15
v = 6.492 m/s
at point B

the normal force is,
F = (m v^2 / r) - m g
0.660 N = (0.0475 kg v^2 / 0.600 m) -( 0.0475 kg)(9.81 m/s2)
0.660 N= 0.0791 v^2 - 0.465975
0.06667 v^2 = 1.125975
v^2 = 16.88878
v = 4.1095 m/s
energy at point A

ground potentential is, GPE and kinetic energy,KE
GPE + KE = m g h + 1/2 m v^2

                 = 0 + 1/2 (0.0475 kg) ((6.492 m/s)^2)

                  = 1.00096 J
similarly,

energy at point B
GPE + KE = m g h + 1/2 m v^2

= (0.0475 kg (9.8 m/s2) (1.2 m)) + (1/2 (0.0475 kg) ((4.1095 m/s)^2) )

=0.5586 kg m2/s2+0.401089

=0.95968 J
so that,
the workdone on the block by friction is the difference between two points

wf=1.00096 J - 0.95968 J = 0.04128 J

so 0.041 is the lost to friction

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