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I have this issue:

If you push a 40.0 kg crate at a constant speed of 1.40 m/s across a horizontal floor (µk=0.25 ), at what rate (a) is work being done on the crate by you and (b) is the energy dissipated by the frictional force?

For starters, here is what I have so far:

RN = mg = 40*9.8 = 392 N

Ff = muRN = 0.25*392 = 98 N

KE = 0.5mv^2 = 0.5*40*(1.40)^2 = 39.2 J

At this point, I don't know how to answer part (a). If W = FD , how do I find the work being done without a distance?

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You are asked for the rate at which work is done (i.e. the power), and you are given a speed which is the rate of change of position... – Michael Brown Mar 2 '13 at 5:22
Yes for calculation rate of work done you don't need the displacement – Dimensionless Mar 2 '13 at 5:25
up vote 0 down vote accepted

SO work done by frictional force will be frictional force * distance trvelled per second or per minute in this it will be 137.2 Jules/sec. And work done by you will be the same because you are applying the force only to keep it moving not for accelerating it which in case would have been F=ma. And if you want to calculate the work done in travelling a particular distance then you wil be given the time interval in which you have to calculate it.

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Great! The answer that was given on the homework was 137 Jules. However, there was no rate per minute or per second given in the answer. This is where it got confusing a bit. In other words 98N * 1.4 m/s = 137.2 J/s – Dimitri Topaloglou Mar 2 '13 at 5:35
glad to know it helped you well in you answer per minute or per second was not given because in question it already asked to find the rate which intern mean's anything happening per second or per minute keep it in your mind it will help you further – Dimensionless Mar 2 '13 at 5:38
Great help! I just couldn't relate the answer with a rate for some reason. I'll keep that in mind for next time. – Dimitri Topaloglou Mar 2 '13 at 5:40

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