What is the Average Power of a Sled Being Pulled with a Constant Force?

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The discussion focuses on calculating the average power generated by a sled being pulled with a constant force of 600 N and an acceleration of 0.08 m/s² over one minute. The initial attempt to find displacement used the incorrect formula, leading to a miscalculation of power. After clarification, the correct displacement was determined using the formula y = (1/2)at², resulting in a displacement of 144 meters. The final calculation for power, using the correct displacement, confirmed that the average power is expressed in watts. The conversation emphasizes the importance of using the correct factors in physics calculations.
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Homework Statement



A sled is being pulled along a horizontal surface by a horizontal force F of magnitude 600 N. Starting from rest, the sled speeds up with acceleration 0.08 m/s^2 for 1 minute.

Find the average power P created by force F.


Homework Equations



P = Fs/T

The Attempt at a Solution



So I need to first find displacement of movement. If y = .08x^2, then at 60 seconds y = 288 ... so displacement is 288 right?

The force is 600 N
so since P = Fs/T, then (600*288)/60 should be my answer right? Also would this answer be in watts? I know that 1 joule/second = 1 watt.
 
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Apart from a tiny factor of 1/2, everything else is right. Yes, the SI unit of power is the Watt. Can you find where you missed the factor? :wink:
 
neutrino said:
Apart from a tiny factor of 1/2, everything else is right. Yes, the SI unit of power is the Watt. Can you find where you missed the factor? :wink:

So y = (1/2).08x^2

this would yield a displacement of 144
so then (600*144)/60

would be a better answer?

Also this answer is in joules so I need to multiply the whole thing by 60 to yield watts right?

so my final answer in watts should be 600*144 ?
 
ssb said:
So y = (1/2).08x^2

this would yield a displacement of 144
so then (600*144)/60

would be a better answer?
That is the correct answer.

Also this answer is in joules so I need to multiply the whole thing by 60 to yield watts right?

As I said, there was nothing wrong with your first answer apart from that 0.5. It is Newtons.metre/second -> Joule/second -> Watt. Moreover, why would you take the trouble of first dividing and then immediately multiplying by 60?
 
neutrino said:
That is the correct answer.



As I said, there was nothing wrong with your first answer apart from that 0.5. It is Newtons.metre/second -> Joule/second -> Watt. Moreover, why would you take the trouble of first dividing and then immediately multiplying by 60?

ack! :blushing: :blushing: :redface:

Thanks buddy! I appreciate the help!
 
is power factor good or bad? Why?

Im just not sure!
 
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