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I'm talking about the maximum speed if let's say I have a car with the power $P = 1000 \text{W}$ and a force of friction of $5 \mbox{N}$ acting in the opposite direction. After some googling I found that the maximum speed is given by $P=Fv$, where $P$ is the power, $F$ is the force, and $v$ is the velocity.

I understand that $W= Fs$ and that $P = W/t$ and $s/t$ is $v$, so yes I understand where the equation comes from, however wouldn't this be the average speed and not the maximum speed? And the force of friction is not the force that's doing the work, so why is it used in the equation?

I hope I've made my question clear enough, thank you in advance!

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  • $\begingroup$ The assumption of constant friction resisting a moving vehicle independent of velocity is completely wrong. $\endgroup$ – Brandon Enright Dec 2 '13 at 19:03
  • $\begingroup$ I apologize for such a mistake, I was just trying to get a very simple example to understand the power to speed relation. $\endgroup$ – user1949350 Dec 2 '13 at 19:13
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  1. The expression $P=Fv$ expresses a relation between the instantaneous power, the force and the velocity. You don't have to average for it to be true.
  2. In your case, the velocity in constant. This implies that the net force is zero. Hence, the force propelling the car is equal and opposite to the friction force. We can then use the magnitude of the friction force.
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  • $\begingroup$ -1: "Power is an instantaneous concept: there is no question of averaging." That's misleading; average power can, in fact, be defined and is a very useful concept. Did you mean to say something like "$P=Fv$ is an expression that holds between the instantaneous power, the force, and the velocity"? $\endgroup$ – joshphysics Dec 2 '13 at 19:15
  • $\begingroup$ Now I'm confused... "P=Fv is an expression that holds between the instantaneous power, the force, and the velocity" Can you please elaborate this a bit more? $\endgroup$ – user1949350 Dec 2 '13 at 19:24
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    $\begingroup$ @joshphysics fair enough, I'll edit it. Just for the record though, I am of course well aware of the concept of average power, I just thought it was a good idea to make it explicit and clear that it is not relevant in this case. $\endgroup$ – Danu Dec 2 '13 at 19:54
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Consider your simple example: a car with a fixed power output of $1000 \text{ W}$ accelerating against a constant frictional force of $5\text{ Newtons }$ (Even better, assume that there is no friction, but the car is climbing a very gentle slope, such that gravity exerts $5\text{ Newtons}$ of force back along the slope.)

Assume that it has reached a stable top speed, $V\text{ m/s}$. In one second, the car will travel $V\text{metres}$. In doing this, it must exert a constant force of $5\text{ Newtons}$, since there is no acceleration, and thus the friction force must be exactly balanced. In doing this, exerting this force through this distance, the car does $(V\times 5) \text{ joules}$ of work.

But in $1$ second, at a power output of $1000 \text{ W}$, the car produces $1000\text{ joules}$ of energy. This energy goes into driving the car, so $$(V\times 5) \text{ joules}=1000\text{ joules}$$More generally:$$W=F\times d$$Dividing by $t$, time:$$\frac{W}{t}=\frac{F\times d}{t}$$or:$$P=F\times v$$

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Terminal velocity ( which is constant ) is where drive power = friction power.

So : 1000 w = friction force * velocity

Then transpose for velocity : velocity = 1000 / 5

= 200 m/s

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