Question

A small block with mass 0.0350 kgkg slides in a vertical circle of radius 0.550 mm 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 AA, the magnitude of the normal force exerted on the block by the track has magnitude 3.90 NN . In this same revolution, when the block reaches the top of its path, point BB, the magnitude of the normal force exerted on the block has magnitude 0.670 NN . Part A How much work was done on the block by friction during the motion of the block from point AA to point BB

Answer 1

Answer:

 W_net = μ 5.58,   μ = 0.1     W_net = 0.558 J

Explanation:

The work is defined by the related

          W = F. d ​​= F d cos θ

where bold indicates vectors.

In the case, the work of the friction force on a circular surface is requested.

The expression for the friction force is

              fr = μ N

the friction force opposes the movement, therefore the angle is 180º and the cos 180 = -1

               W = - fr d

the path traveled half the length of the circle

               L = 2 π R

               d = L / 2

               d = π R

we substitute

                 W = - μ N d

Total work is initial to

                W_neto = - μ π R (N_b - N_a)

let's calculate

                W_net = - μ π 0.550 (0.670 - 3.90)

                W_net = μ 5.58

for the complete calculation it is necessary to know the friction coefficient, if we assume that μ = 0.1

                W_net = 0.1 5.58

                W_net = 0.558 J