The net force acting on a 13.0-kg object is given by F_x = (20 − x)N,where F_x is in newtons and x is in meters.How much work is done on the object as it moves from x = 0 to x = 13 m? A) 7 J B) 91 J C) 175.5 J D) 149 J E) 87.75 J

Question

Quizplus · Accepted Answer

The work done on an object is equal to the area under the force vs. displacement curve. In this case, the force varies with displacement according to F_x = (20-x)N. Integrating this from x=0 to x=13 gives: W = ∫(20-x)dx from x=0 to x=13 W = [20x - (x^2)/2] from x=0 to x=13 W = [20(13) - (13^2)/2] - [20(0) - (0^2)/2] W = 175.5 J Therefore, the answer is C, 175.5 J.

The Net Force Acting on a 13