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

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Homework Help Overview

The problem involves calculating the average power exerted by a constant force on a sled being pulled along a horizontal surface. The sled starts from rest and accelerates at a specified rate over a given time period.

Discussion Character

  • Mathematical reasoning, Problem interpretation, Assumption checking

Approaches and Questions Raised

  • Participants discuss the calculation of displacement using the formula for motion under constant acceleration and question the application of the power formula. There is an exploration of the impact of a missing factor in the calculations.

Discussion Status

The discussion is ongoing with participants examining the calculations and clarifying the correct application of formulas. Some guidance has been provided regarding the correct displacement calculation and the relationship between joules and watts.

Contextual Notes

Participants are working under the constraints of a homework assignment and are questioning the assumptions made in their calculations, particularly regarding the displacement and the application of the power formula.

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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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