0
$\begingroup$

I've come across a textbook question and I've got myself confused by it although it seems like it should be simple. The van accelerates with a (supposedly) constant resistance force up to its top speed. Obviously at top speed the force from the engine is the same as the resistive force. In the real world the resistive force would increase with speed, but why is it the case that the force from the engine will decrease as the speed increases? The power from the engine is constant, so why isn't the force?

enter image description here

$\endgroup$
  • $\begingroup$ Are you familiar with work = force * distance? $\endgroup$ – BowlOfRed Jan 28 at 23:27
  • $\begingroup$ The force from the engine decreases with speed for a constant power output because an alternate form of the power equation is $P=Fv$. $\endgroup$ – David White Jan 29 at 0:53
  • $\begingroup$ @David White But does the derivation of $P = Fv$ not rely on a constant force when you take the derivative of work done, leaving $F \times \frac{dx}{dt}$? $\endgroup$ – TIF Jan 29 at 15:34
  • $\begingroup$ @TIF, no. I'll make the derivation an answer. See below. $\endgroup$ – David White Jan 29 at 16:26
0
$\begingroup$

The power equation is typically written as $P=W/t$. This equation is not easily used when an object such as an automobile is accelerating under constant power, and one wants to know the force involved in that acceleration as a function of time. Accordingly, this work-time equation can be manipulated in order to derive another equation for power that includes force and velocity:

$P=W/t$

$W=Fd$

A substitution of the work equation into the power equation yields:

$P=Fd/t$

$P=F(d/t)$

Since $d/t$ is equal to velocity,

$P=Fv$

Thus, it is easily seen that under a constant power condition, force decreases as velocity increases, such that

$F=P/v$

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.