A horizontal force of 200 N is applied to a 55.0-kg cart across a 15.0-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)-4.7E+3 J B)-1.4E+3 J C)-3.0E+3 J D)-9.0E+1 J E)-3.1E+2 J

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Quizplus · Accepted Answer

The net force on the cart is given by F_net = mass * acceleration = 55 kg * 2 m/s² = 110 N. The applied force is 200 N, so the force of friction is 200 N - 110 N = 90 N. The work done by friction is W = force * distance * cos(θ), where θ = 180° since friction opposes motion, so W = 90 N * 15 m * (-1) = -1350 J, which is approximately -1.4E+3 J.

A Horizontal Force of 200 N Is Applied to a 55.0-Kg