Physics Stack Exchange is a question and answer site for active researchers, academics and students of physics. It's 100% free.

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

I'm a pure mathematician by trade, but have been teaching myself classical mechanics. I've got to the chapter on Work, Energy and Power and I've found an example that is causing me some problems.


A van of mass 1250 kg is travelling along a horizontal road. The van's engine is working at 24 kW. The constant force of resistance to motion has magnitude 600 N. Calculate:

  1. The acceleration of the van when it is travelling at 6 m/s.
  2. The maximum speed of the van.


Part 1 is fine. We know that $P = Fv$ and so $24000 = 6F$. It follows that $F = 4000$ N. To find the acceleration, we use $F=ma.$ The total resulting force is the traction minus the resistance, in other words $F = 4000 - 600 = 3400$ N. The mass is 1250 kg and so $3400 = 1250a$. It now follows that $a=2.72$ m/s/s.

Part 2 is the part that causes me problems. The maths isn't the problem, it's the way it's used. At maximum speed the acceleration is zero and so the resultant horizontal force will be zero. The book says that $T' = 600$. There are two problems here: I'm used to a prime denoting differentiation. I guess it just means the new tractive force. But the resistance force is working against the direction of motion, so shouldn't we have $T' = -600$?

Running with $T' = 600$, the book then uses $P = Fv$ to get $24000=600v$ and so $v=40$ m/s. I can see that using $T'=-600$ would give a negative velocity, which is clearly untrue.

In short

I don't see why the fact that at maximum speed there will be no acceleration, so the resultant horizontal force will be zero leads to us using $T'=600$ when $T > 0$ was the forwards tractive force and $-600 < 0$ was the resistance force.

share|cite|improve this question
up vote 2 down vote accepted

I don't think you initially defined $T$, so I'm not entirely sure what it originally used for. The overall idea, however, is that because the newt force is zero (and the resistive force stays the same, i.e. $-600$, then the new forward force has to be $T' = +600$.

Physics is often (e.g. besides high-energy) much more pragmatic than mathematics per se -- one expression of this is the mixed use of things like primes, or superscripts, etc. Similarly, negative signs often aren't tracked quite as closely (as they should be). Because the positive or negative are purely conventional, and it is conceptually clear that the resulting force (and velocity) need to be opposite the resistive force -- then you can just give them the appropriate sign. It is, of course, better to treat them accurately and consistently, but note that this is a common location of hand-waving in introductory level textbooks.

share|cite|improve this answer
Sorry: $T$ is used for the tractive force, i.e. the total pulling power of the vehicle. – Fly by Night Dec 26 '12 at 22:59
I have noticed quite a lot of notation misuses in the textbook I've been reading. For example they use $F$ for a force and $F$ for the magnitude of that force. I would use ${\bf F}$ and $||{\bf F}||$ respectively. I think I have resolved the problem. As you say, the net residence is $-600$ N and so the engine must produce a net force of $+600$ N in order to maintain constant speed. That is the force used in the calculations. I can understand why students struggle so much: some of the textbooks leave a lot unsaid. Thanks a lot for taking the time to reply, it's much appreciated – Fly by Night Dec 26 '12 at 23:00
Happy to help!! – DilithiumMatrix Dec 27 '12 at 21:44

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

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