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timofeeve [1]
3 years ago
13

A cart for hauling ore out of a gold mine has a mass of 413 kg, including its load. the cart runs along a straight stretch of tr

ack that is sloped 4.69° 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 4.10 × 102-n force for a distance of 175 m by means of a rope that is parallel to the ground and makes an angle of 14.3° with the track. the coefficient of friction for the cart\'s wheels on the track is 0.0163. use g = 9.81 m/s2. find the work that the donkey performs on the cart during this process.
Physics
1 answer:
Andrei [34K]3 years ago
8 0
Given from the problem :mass m = 413 kg;coefficient of friction u = 0.0163;acceleration due to gravity g = 9.8 m/s2;inclined angle @1 = 14.3;inclined angle  @2 = 4.69;distance travelled d = 175 m;applied fore F = 410 N;  the component of the force from the donkey in the direction of motion isF2 = F1 [email protected]= 397.2964498768165 N
Fy = N - mg [email protected] = 0N = mg [email protected] = 4037.964151113007 NFx = F2 - mg [email protected] - f = mahere f = u N=65.8188156631420141

F2 - mg [email protected] - f = maa = F2 - mg [email protected] - f/ m=0.31923412183075155 m/s^2
work done by donkeyW = F2 d=69526.8787284428875 J
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An 81.5-kg man stands on a horizontal surface.
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a)V= 0.0827 m³

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Suppose the maximum power delivered by a car's engine results in a force of 16000 N on the car by the road. In the absence of an
joja [24]

Answer:

Approximately 9.7\; \rm m \cdot s^{-2}.

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Assuming that there is no other force on this vehicle, the 16000\; \rm N force from the road would be the only force on this vehicle. The net force would then be equal to this 16000\; \rm N\! force. The size of the net force would be 16000\; \rm N\!\!.

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By Newton's Second Law of motion, the acceleration of this vehicle would be proportional to the net force on this vehicle. In other words, the acceleration of this vehicle, a, would be:

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For this vehicle, \Sigma F = 16000\; \rm N whereas m = 1650\; \rm kg. The acceleration of this vehicle would be:

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3 years ago
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