Question

A sled of mass 82.0 kg starts from rest and slides down a 11.5° incline 94.0 m long. It then travels for 20.0 m horizontally before starting back up an 7.5° incline. It travels 94.0 m along this incline before coming to rest. What is the magnitude of the net work done on the sled by friction?

Answer 1

The magnitude of the net work done on the sled by friction is 2.62 x 10^3 J.

Step-by-step explanation:

Part 1: Work Done by Gravity on the First Incline

Mass (m): 82.0 kg

Gravitational acceleration (g): 9.81 m/s^2

Weight (mg): 82.0 kg * 9.81 m/s^2 = 806.42 N

Angle of incline (θ): 11.5°

Height of incline (h): sin(θ) * 94.0 m = sin(11.5°) * 94.0 m ≈ 17.47 m

Work done by gravity (W_g): mgh = 806.42 N * 17.47 m ≈ 14123 J

Part 2: Work Done by Friction on the First Incline

Energy conservation: Work done by friction is the opposite of work done by gravity.

Work done by friction (W_f1): W_f1 = -W_g = -14123 J

Part 3: Work Done by Friction on the Horizontal Path

Horizontal component of weight (F_h): mgcos(θ) = 806.42 N * cos(11.5°) ≈ 794.07 N

Horizontal distance traveled (d): 20.0 m

Work done by friction (W_f2): F_h * d = 794.07 N * 20.0 m ≈ 15881.4 J

Part 4: Work Done by Gravity on the Second Incline

Angle of incline (θ): 7.5°

Height of incline (h): sin(θ) * 94.0 m = sin(7.5°) * 94.0 m ≈ 11.35 m

Work done by gravity (W_g): mgh = 806.42 N * 11.35 m ≈ 9156.57 J

Part 5: Work Done by Friction on the Second Incline

Energy conservation: Work done by friction is the opposite of work done by gravity.

Work done by friction (W_f3): W_f3 = -W_g = -9156.57 J

Part 6: Net Work Done by Friction

Total work done by friction: W_net = W_f1 + W_f2 + W_f3

W_net: -14123 J + 15881.4 J - 9156.57 J ≈ 2621.83 J

Magnitude of net work: |W_net| = 2621.83 J ≈ 2.62 x 10^3 J