Homework Help: General Physics I

Posted by VitaX on Wednesday, October 13, 2010 at 4:08am.

A 50-N force acts on a 2-kg crate that starts from rest. When the force has been acting for 2 s the rate at which it is doing work is:

I found this question online so I saw the answer is 2500 Watts. What I don't understand is how to come to this conclusion. Power = Work/time. The time is known but to find Work we need both the force and distance. How do I go about finding them? I assume I have to use that mass value somewhere in here, just not sure where.

General Physics I - VitaX, Wednesday, October 13, 2010 at 5:58am
I was able to figure some things out. Like a = F/m = 50/2 = 25 m/s^2
s=Vo +.5at^2 = .5*25*2^2 = 50 m
Work = 50*50 = 2500 J
Power = Work/time = 2500/2 = 1250 W

Problem is that the answer is 2500, but the multiple choice answers are all in Watts so that has to mean Power. What did I do wrong because either way I look at it, the answer is 1250 W for Power.
General Physics I - drwls, Wednesday, October 13, 2010 at 6:40am
It is a pleasure to help you with this; thank you for showing your work and thought process. Thank you for using Jiskha

instantaneous Power delivered by a moving force equals:
P = force x velocity

The final velocity in this case is
Vf = (1/m)*(final momentum) = F*t/m
= 50*2/2 = 50 m/s

Power at end of interval = F*Vf = 50*50 = 2500 W

The reason you got a different number is that you were calculating the AVERAGE power during the 2 second interval. The average is half the final value while the acceleration continues.

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