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?

w = -0.475N

Explanation:

To get Va and Vb

R = 0.525 m

m = 0.0350 kg

g = 9.8 m/s²

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

b

explanation:

initial speed when it is dropped is 150 ft/s

now at the same moment the acceleration due to gravity is perpendicular to its velocity

so as we know that

now we have centripetal acceleration a = 32 ft/s^2 as its perpendicular to velocity

while tangential acceleration at the same moment will be zero

so

now radius of curvature is given as

so the radius of curvature just after it is dropped is 703.125 ft