A force of 120 newtons (N) was applied to a pulley to lift an object a height of 1 meter (m). If 100 joules (J) of work was done on the weight, how much work was used to overcome the force of friction in the pulley?

A. 1.2 joules

B. 20 joules

C. 120 joules

D. 220 joules

1)   F = 8.789 10² N, 2)   F = 1.5 10⁴ N

Explanation:

We can solve this exercise using Coulomb's law

         F =

Knowing that charges of the same sign repel

let's apply this equation to our case

1) the charges are q = - 0.0025 C and the distance between them r = 8 m

       we calculate

           F = 9 10⁹ 0.0025 0.0025 /8²

           F = 8.789 10² N

as the two charges are of the same sign the force is repulsive

2) q₁ = -0.004C and q₂ = -0.003 C with a distance of r = 3.0 m

    we calculate

             F = 9 10⁹ 0.004 0.003 / 3²

             F = 1.5 10⁴ N