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

  • Thread starter Thread starter ssb
  • Start date Start date
  • Tags Tags
    Average Power
Join the discussion
Registration is free. Start your own thread to ask a follow-up.
5 replies · 29K views
ssb
Messages
119
Reaction score
0

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.
 
Physics news on Phys.org
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!