A cart for hauling ore out of a gold mine has a mass of 401 kg, including its lo

Question

A cart for hauling ore out of a gold mine has a mass of 401 kg, including its load. The cart runs along a straight stretch of track that is sloped 4.85 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 413-N force for a distance of 153 m by means of a rope that is parallel to the ground and makes an angle of 13.5 degree with the track. The coefficient of friction for the cart's wheels on the track is 0.0175. 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

Tension in the string be T
Normal reaction be N
Angle of incline be theta
Angle the rope makes with the track be phi
Fricion be f

so, Tcos(phi) - mgsin(theta) - f = ma
and, mgcos(theta) = N
and, f = mu*N
so, f = mu*mgcos(phi)
and, Tcos(phi) - mgsin(theta) - mu(mgcos(phi)) = ma

m = 401kg
theta = 4.85
phi = 13.5
mu = 0.0175
g = 9.81
T = 413N
so
401.589 - 332.593 - 66.94 = 401*a
a = 0.00512 m/s^2

Now.
distance covered by the trolley along the incline, d = 153m
Work done by friction = fd = 66.94*153 = 10,241.82 J
Work done by gravity = mg*d*sin(theta) = 50,886.801 J
Velocity of cart after 153m, v = sqroot(2*a*d) = 1.2516 m/s
K.E. of cart = 314.127 J
Work done by mule = Sum of all works = 61442.748 J

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