The net force acting on a 4.92-kg object is given by F_x = (10 − 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 = 12.4 m? A)61.0 J B)12.4 J C)47.1 J D)23.6 J E)35.3 J

Question

Quizplus · Accepted Answer

To calculate the work done, we need to integrate the force over the given displacement.

∫W = ∫F dx = ∫(10 - x) dx

∫W = [10x - (1/2)x^2] from x = 0 to x = 12.4

∫W = [10(12.4) - (1/2)(12.4)^2] - [10(0) - (1/2)(0)^2]

∫W = 124 - 0 = 124 J

Therefore, the work done on the object as it moves from x = 0 to x = 12.4 m is 47.1 J (closest to the answer choice C).

The Net Force Acting on a 4