Question

A small block with mass 0.0400 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 4.05 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 0.685 N .

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

Answer #1

here,

the mass of small block , m = 0.04 kg

the radius of vertical circle , r = 0.6 m

let the speed at the bottom of loop be vB

equating the forces vertically

normal force , N = m * ( g + vb^2 /r)

4.05 = 0.04 * ( 9.81 + vB^2 /0.6)

solving for vB

vB = 7.41 m/s

let the speed at the top of loop be vt

equating the forces vertically

normal force , N = m * ( g - vt^2 /r)

0.685 = 0.04 * ( 9.81 + vt^2 /0.6)

solving for vt

vt = 2.09 m/s

using Work energy theorm

the work done by friction , W = (0.5 * m * vt^2 + m * g * (2r)) - 0.5 * m * vB^2

W = (0.5 * 0.04 * 2.09^2 + 0.04 * 9.81 * 2 * 0.6) - ( 0.5 * 0.04 * 7.41^2) J

W = - 0.54 J

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