The net force acting on a 12.6-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 = 10 m? A)300 J B)200 J C)150 J D)100 J

Question

Quizplus · Accepted Answer

The work done on an object is given by the formula:

W = ∫Fdx

where F is the force acting on the object and x is the distance moved by the object.
So, we need to integrate the given force expression with respect to x from 0 to 10 m.

W = ∫(20 - x) dx from x = 0 to x = 10
W = [20x - (x^2)/2] from x = 0 to x = 10
W = [20(10) - (10^2)/2] - [20(0) - (0^2)/2]
W = 200 - 0
W = 200 J

Therefore, the work done on the object as it moves from x = 0 to x = 10 m is 200 J. Hence, the best choice is C) 150 J.

The Net Force Acting on a 12