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Applying W=F*s for variable force

  • #1

Homework Statement


A car of mass 2000kg moves along a horizontal road against a constant resistance of manitude (P)N. The total work done by the engine in increasing its speed from 4ms^-1 to 5.5ms^-1 while it moves a distance of 60m is 30000J. Find P.

Homework Equations


ΔEk+WP=WE

The Attempt at a Solution


Straightforward question. The correct solution is as follows:
ΔEk+WP=WE

1/2(2000)(5.5^2-4.0^2)+60P=30000
P=262.5J

However, another solution was proposed:

The work done by the engine is 30000J and the distance it moved was 60m so the average force it exerted was 30000/60=500N

500-P=ma
finding a using v^2=u^2+2as, and then multiplying it by 2000 (m) gives ma=237.5
P=262.5N

This alternative solution gives the same answer. However, a classmate pointed out that it is incorrect because he said it assumed a constant driving force, which is a wrong assumption. I think that dividing the total work done (30000J) by the total distance moved will give the AVERAGE value of this varying driving force and the value of a is the is the AVERAGE value of this varying acceleration. Is there a flaw in this method? If so, what is it?

Thanks in advance
 

Answers and Replies

  • #2
haruspex
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it is incorrect because he said it assumed a constant driving force,
That is a valid criticism of the method, but it turns out not to matter in this case.
Indeed, v2=u2+2as is one of the SUVAT equations, and as a set those are only supposed to be for constant acceleration, i.e. the same assumption. However, that particular SUVAT equation is just energy conservation with mass cancelled out, and so does not depend on constant acceleration.
The cleanest method would therefore be to use energy conservation rather than SUVAT.
 
  • #3
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Your force balance equation is $$F-P=m\frac{dv}{dt}$$. If we multiply both sides of this equation by ##\frac{ds}{dt}=v##, we obtain:
$$F\frac{ds}{dt}-P\frac{ds}{dt}=mv\frac{dv}{dt}=\frac{m}{2}\frac{dv^2}{dt}$$If we next integrate this equation between 0 and t, we obtain:$$\int_0^s{Fds}-Ps=m\frac{v^2(t)-v^2(0)}{2}$$Dividing both sides of this equation by s yields:$$\frac{1}{s}\left[\int_0^s{Fds}\right]-P=m\frac{v^2(t)-v^2(0)}{2s}$$If we define the average force as $$\bar{F}=\frac{1}{s}\left[\int_0^s{Fds}\right]$$and the average acceleration as $$\bar{a}=\frac{v^2(t)-v^2(0)}{2s}$$we obtain:$$\bar{F}-P=m\bar{a}$$But this interpretation depends strictly on defining the average force and the average acceleration in this very specific way.
 

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