A horizontal force of 200 N is applied to a 55-kg cart across a 10-m level surface.If the cart accelerates at 2.0 m/s²,then what is the work done by the force of friction as it acts to retard the motion of the cart? A)-1 100 J B)-900 J C)-800 J D)-700 J

Question

Quizplus · Accepted Answer

First, we need to find the net force acting on the cart:

F_net = F_applied - F_friction
F_net = 200 N - F_friction

Using Newton's Second Law, we can find the acceleration of the cart:

F_net = ma
200 N - F_friction = (55 kg)(2.0 m/s^2)
F_friction = 90 N

Now, we can use the work-energy principle to find the work done by the force of friction:

W_friction = -ΔK
W_friction = -(1/2)(m)(v_f^2 - v_i^2)
W_friction = -(1/2)(55 kg)((0 m/s)^2 - (2.0 m/s)^2)
W_friction = -110 J

Since the force of friction acts in the opposite direction of motion, the work done is negative. Therefore, the best choice is B) -900 J.

A Horizontal Force of 200 N Is Applied to a 55-Kg