What is the average power output of the engine of a truck when the truck travels up a 6km long hill out of an open pit mine? The truck with the load, has a mass of 160,000kg, and starts from rest at the botttom of the hill. When the truck took 42 seconds to reach the top of the hill. The average drag force on the truck during the time it is moving is 200N

First figure the work the truck does:

1. The truck lifts itself out of the mine. For a truck of mass m, the work to do this W = m*g*h, where h is the vertical climb. You don't tell us what that value is - I assume you have a diagram or something that you can get this from, right?

2. The truck does work overcoming drag: W = force * distance, or 200N * 6 Km.

3. Finally, given that the truck accelerates from rest, the work to overcome the inertia of the truck is W = m*a*d, where a is the acceleration of the truck and d is the distance over which it acceerates. You don't give enough information to determine what the value of a is - did you leave something out?

Add these work elements togeter to get the total work done by the truck. Then divide that work by the time it took to do it (presumably 42 seconds) to get the average power output. You'll have to do some messing with the units to get it to come out right.

One last thing - if the truck indeed travels 6 Km in only 42 seconds it is one fantastic truck! Are you sure you have the right data here?

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Originally Posted by ebaines

One last thing - if the truck indeed travels 6 Km in only 42 seconds it is one fantastic truck!

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True, but applause for the driver : going uphill with 160 tons at some 514 Km/hr.
But even it that were 42 minutes and 8,6 Km/hr, I still take my cap off to him or her ! :)