Question

A small block with mass 0.0350kg slides in a vertical circle of radius 0.525m 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 3.85N 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 0.665N .
How much work was done on the block by friction during the motion of the block from pointA to point B?

Answer 1

Answer:

w = -0.475N

Explanation:

$K.E_{a} + P.E_{a} + W_{fr} = K.E_{b} + P.E_{b}\\K.E = 0.5mv^{2} \\Normal force at point B, N_{B} = 0.665N\\Normal force at point A, N_{A} = 3.85N$

To get Va and Vb

$F = mv_{A} ^{2} /R................(1)\\F = N_{A} - mg.........................(2)\\mv_{A} ^{2} /R = N_{A} - mg\\v_{A} ^{2} = R (N_{A}/m - g)\\v_{A} = \sqrt{ R (N_{A}/m - g)}$

R = 0.525 m

m = 0.0350 kg

g = 9.8 m/s²

$v_{A}= \sqrt{0.525(3.85 /0.0350 - 9.8)} \\v_{A} = 7.25 m/s$

K.Ea = 0.5 * 0.035 * 7.25²

K.Ea = 0.92 J

Since point A is at the bottom of the path, h = 0 m

P.Ea = 0 m

For Vb

$F = mv_{B} ^{2} /R................(1)\\F = N_{B} - mg.........................(2)\\mv_{B} ^{2} /R = N_{B} - mg\\v_{B} ^{2} = R (N_{B}/m - g)\\v_{B} = \sqrt{ R (N_{B}/m - g)}$

$N_{B} = 0.665N$

$v_{B}= \sqrt{0.525(0.665 /0.0350 - 9.8)} \\v_{B} = 2.198 m/s$

$K.E_{B} = 0.5* 0.035 * 2.198^{2} \\K.E_{B} = 0.085 J$

$P.E_{B} = mgh_{B} \\h_{B} = A diameter = 2R = 2 * 0.525\\h_{B} = 1.05 m\\P.E_{B} = 0.035 * 9.8 * 1.05\\P.E_{B} =0.36 J$

$from K.E_{a} + P.E_{a} + W_{fr} = K.E_{b} + P.E_{b}$

$0.92 + w_{fr} + 0 = 0.085 + 0.36\\ w_{fr} = -0.475J$