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