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A cart for hauling ore out of a gold mine has a mass of 435 kg, including its lo

ID: 1343172 • Letter: A

Question

A cart for hauling ore out of a gold mine has a mass of 435 kg, including its load. The cart runs along a straight stretch of track that is sloped 4.93° 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 449-N force for a distance of 173 m by means of a rope that is parallel to the ground and makes an angle of 13.3° with the track. The coefficient of friction for the cart's wheels on the track is 0.0165. Use g = 9.81 m/s2. Find the work that the donkey performs on the cart during this process.

B-Find the work that the force of gravity performs on the cart during the process.

G- Calculate the work done on the cart during the process by friction.

Explanation / Answer

here,

mass of the cart , m = 435 kg

theta = 4.93 degree

force exerted by the donkey , fd = 449 N

distance , d = 173 m

phi = 13.3 degree

uk = 0.0165

the work done by the donkey , Wd = fd * d* cos(phi)

Wd = 449 * 173 *cos(13.3)

Wd = 7.56 * 10^4 J

the work that the donkey performs on the cart during this process is 7.56 * 10^4 J

(B)

the work that the force of gravity performs on the cart during the process , Wg = m*g*d*sin(theta)

Wg = 435 * 9.8* 173 * sin(4.93)

Wg = 6.34 * 10^4 J

the work that the force of gravity performs on the cart during the process is 6.34 * 10^4 J

(C)

the work done on the cart during the process by friction.

Wf = uk*d *( m*g*cos(theta) - fd*sin(phi))

Wf = 0.0165 *173* ( 435 * 9.8* cos(4.93) - 449 * sin( 13.3))

Wf = 1.18 * 10^4 J

the work done on the cart during the process by friction is 1.18 * 10^4 J

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