A cart for hauling ore out of a gold mine has a mass of 405 kg, including its lo
ID: 1642530 • Letter: A
Question
A cart for hauling ore out of a gold mine has a mass of 405 kg, including its load. The cart runs along a straight stretch of track that is sloped 4.65° from the horizontal. A donkey, trudging along and to the side of the track, has the unenviable job of pulling the cart up the slope with a 404-N force for a distance of 151 m by means of a rope that is parallel to the ground and makes an angle of 13.7° with the track. The coefficient of friction for the cart's wheels on the track is 0.0177. Use g = 9.81 m/s2. Find the work that the donkey performs on the cart during this process. _____J Find the work that the force of gravity performs on the cart during the process. ______J Calculate the work done on the cart during the process by friction. _____J
Explanation / Answer
The work done by gravity is equal to the change in potential energy:
W done by gravity = -m*g*h
It's negative because gravity pulls in the opposite direction to the cart's displacement.
h = (151 m) * sin(4.65°)
W done by gravity = -(405 kg)*(9.81 m/s^2)*(151 m) *sin(4.65°)
W done by gravity = -4.86 * 10^4 J
W done by donkey = (151 m) * (404 N) *cos(13.7°)
W done by donkey = 5.927 * 10^4 J
Work done by friction:
F = -0.0177 * mg*cos(4.65°)
The negative sign is because force is in the opposite direction of displacement.
W done by friction = F * (151 m)
W done by friction = -0.0177 * mg*cos(4.65°) * (151 m)
W done by friction = -0.0177 * (405 kg) * (9.81 m/s^2) *cos(4.65°) * (151 m)
W done by friction = -1.058 * 10^4 J
Related Questions
drjack9650@gmail.com
Navigate
Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.