Question

A 16 kg block is dragged over a rough, hor izontal surface by a constant force of 160 n acting at an angle of 28.1 ◦ above the horizon tal. the block is displaced 14.8 m, and the coefficient of kinetic friction is 0.129. 16 kg µ = 0.129 160 n 28.1 ◦ find the work done by the 160 n force. the acceleration of gravity is 9.8 m/s 2 . answer in units of j.

Answer 1

The work done by the 160 N force is 2038.95 J.

Step-by-step explanation:

To find the work done by the 160 N force, we need to calculate the component of the force in the direction of displacement. Using the given angle, we can find the vertical component of force. Fv = F * sin(θ) = 160 N * sin(28.1°) = 72.95 N. Since the block is displaced horizontally and the friction force acts in the opposite direction, there is no vertical displacement. Therefore, the vertical component of the force does no work. Now we can find the horizontal component of the force. Fh = F * cos(θ) = 160 N * cos(28.1°) = 137.49 N. The horizontal component of the force does work, so the work done by the 160 N force is given by W = Fh * d = 137.49 N * 14.8 m = 2038.95 J. Therefore, the work done by the 160 N force is 2038.95 J.