Question

Ok so I'm lost, my teacher said that the answers I got for a though c are correct but I don't know how I would go about doing d. He told me that I should get 42.9m. Can someone explain this to me and show me what the formula to solve for this would be used. Thank You!

1. A 2.63 kg object moves 56.8 m on a horizontal surface whose coefficient of,kinetic friction is 0.258. While it moves, a 42.2 N force acts on it at an angle of 33.8° with respect to the direction the object moves in. a) Identify all the gorces that act on the object while it moves FK b) Calculate the work done by each force identified in part (a). c) Calculate the net work done on the object d) How far would the object have to move, with the same forces acting on it, for the net work to be 1.22 x 103J?

Answer 1

(d) Net work done by force = 1.22 times 10^3 J work by Applied force + work done by friction = 1.22 times 10^3 J (42.2 times cos(33.8) times d) + work by friction = 1220 equation (1) Friction force = mu N N = (mg - 42.2 sin(33.8)) 35.07 d - (mu) d = 1220 35.07d - 6.65d = 1220 d = 1220/28.43 d = 42.91 meter