Question

012 (part 1 of 2) 10.0 points

A 19.4 kg block is dragged over a rough, hor- izontal surface by a constant force of 80.4 N acting at an angle of 28.7◦ above the horizon- tal. The block is displaced 61.8 m, and the coefficient of kinetic friction is 0.113.
19.4 kg μ = 0.113
Find the work done by the 80.4 N force. The acceleration of gravity is 9.8 m/s2 .
Answer in units of J.
013 (part 2 of 2) 10.0 points
Find the magnitude of the work done by the force of friction.
Answer in units of J.

Answer 1

The work done by the applied force (W_applied) is approximately 4484.074 J, and the work done by the force of friction (W_friction) is approximately -1331.874 J. The negative sign for W_friction indicates that the force of friction acts opposite to the direction of displacement.

Work done by the applied force (W_applied):

W_applied = F_applied * d * cos(theta)

W_applied = 80.4 N * 61.8 m * cos(28.7 degrees)

W_applied = 80.4 N * 61.8 m * 0.882

W_applied is approximately 4484.074 J

Work done by the force of friction (W_friction):

N = m * g

N = 19.4 kg * 9.8 m/s^2

N is approximately 190.12 N

W_friction = -mu * N * d

W_friction = -0.113 * 190.12 N * 61.8 m

W_friction is approximately -1331.874 J