A cart for hauling ore out of a gold mine has a mass of 427 kg, including its lo

Question

A cart for hauling ore out of a gold mine has a mass of 427 kg, including its load. The cart runs along a straight stretch of track that is sloped 4.93 degree 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 4.50 times 10^2-N force for a distance of 143 m by means of a rope that is parallel to the ground and makes an angle of 12.7 degree with the track. The coefficient of friction for the cart's wheels on the track is 0.0189. Use g = 9.81 m/s^2. Find the work that the donkey performs on the cart during this process. Find the work that the force of gravity performs on the cart during the process. Calculate the work done on the cart during the process by friction.

Explanation / Answer

given:
mass=427kg
slope = 4.93 degrees
force = 4.50*10^2N = 450 N
distance =143m
angle = 12.7 degree
coefficient of friction = 0.0189

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.
where h = (143 m) * sin(4.93)
W done by gravity =- 427*9.81*143*sin4.93 = -51477.94 J

Work done by they donkey equals the distance moved multiplied by the component of the force applied in the direction of movement:
W done by donkey = (143) * (450) *cos(12.7)
W done by donkey= 62775.94 J

Work done by friction:
Energy is conserved, so we know that the work done by the donkey plus the work done by gravity plus work done by friction must equal zero. You found the work done by gravity and the work done by the donkey, so,

Work done by friction = -Work done by gravity - Work done by donkey
Work done by friction = 51477.94 - 62775.94 = -11298 J

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