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Work and Horsepower

 
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Jun4-06, 08:38 PM   #1
 

Work and Horsepower


A skier of mass 65 kg is pulled up a slope by a motor-driven cable, a)How much work is required to pull him 50 m up a 30° slope (assumed frictionless) at a constant speed of 2.9 m/s? b)how much horsepower must the motor have to preform this task?

Please help!! I don't know what to do with this... I know it's a simple question but i've managed to get all confused. Where do I start? Thank you!
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Jun4-06, 09:54 PM   #2
 
Find the force pulling the skier down the hill. This is the force required to pull him up. That is a good place to start. From there, remember that KE=(1/2)mv^2 and also that W=Fd.
Jun4-06, 09:58 PM   #3
 
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Quote by netprincess
A skier of mass 65 kg is pulled up a slope by a motor-driven cable, a)How much work is required to pull him 50 m up a 30° slope (assumed frictionless) at a constant speed of 2.9 m/s? b)how much horsepower must the motor have to preform this task?
a) How long does it take to cover 50 m if the speed is 2.9 m/sec? How much vertical distance does it rise in that time? How much energy did it use in that time? What is the rate of work done/unit time?

b) How many Joules/sec is 1 horsepower?

AM
Jun4-06, 10:02 PM   #4
 
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Work and Horsepower


Quote by Mr. Snookums
Find the force pulling the skier down the hill. This is the force required to pull him up. That is a good place to start. From there, remember that KE=(1/2)mv^2 and also that W=Fd.
There is no change in kinetic energy in the problem as stated, so the power is not used to provide kinetic energy.

AM
Jun4-06, 10:20 PM   #5
 
Quote by Andrew Mason
There is no change in kinetic energy in the problem as stated, so the power is not used to provide kinetic energy.

AM
If it has velocity and mass, doesn't it have kinetic energy?
Jun4-06, 10:26 PM   #6
 
The skier does have kinetic energy when he's being pulled. I think what Andrew meant was that there was no change in this KE, as the question didn't mention about the skier starting from rest, but being pulled with constant velocity instead.
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