Question

# A small block with mass 0.0350 kg slides in a vertical circle of radius 0.450 m...

A small block with mass 0.0350 kg slides in a vertical circle of radius 0.450 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 3.90 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 0.670 N .

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

if we find the total of GPE and KE at both positions
the difference between these energies will be energy lost to friction (work on block by friction)
take the h at A as 0 and the h at B will be 1.15 m (1 diameter)
now we need the v's at A and B
at A
Fc = m v^2 / r + m g
3.90 = (0.035 v^2 / 0.450) + 0.035(9.8)
3.90 = 0.0777 v^2 + 0.343
v = 6.76 m/s
at B
Fc = m v^2 / r - m g
0.670 = (0.035 v^2 / 0.450) - 0.035(9.8)
0.670 = 0.077 v^2 - 0.343
v = 3.62 m/s
energy at A
GPE + KE = m g h + 0.5 m v^2 = 0 + 0.5 (0.035) (6.76^2) = 0.799 J
energy at B
GPE + KE = m g h + 0.5 m v^2 = (0.035 (9.8) (1.15)) + (0.5 (0.035) (3.62^2) ) = 0.623 J
so
0.799 - 0.623 = 0.176 J lost to friction

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