Question

the diagram below represents a 4.0-newton force applied to a 0.200-kilogram copper block sliding to the right on a horizontal steel table

Answer 1

Question is not complete, so i have attached it.

Answer:

A) W = 1.962 N

B) F_f = 0.706 N

C) F_net = 3.294 N

Explanation:

A) The block is 0.2 kg.

Formula for weight is;

W = mg

W = 0.2 × 9.81

W = 1.962 N

B) Magnitude of the force acting on the moving block is given by;

F_f = μmg

Where μ is coefficient of kinetic friction

Coefficient of kinetic friction for copper(mild steel) from online sources is 0.36

Thus;

F_f = 0.36 × 1.962

F_f = 0.706 N

C) Magnitude of net force is;

F_net = F_applied - F_f

F_net = 4 - 0.706

F_net = 3.294 N