# Finding work and average force

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1. Jun 6, 2015

### Seejianshin

1. The problem statement, all variables and given/known data
A 500hp Mustang is able to push this 1750kg car down a 410m drag strip in a time of 11.5s.
a) What is the work done by the car?
b) What is the average force during the car's motion.
Given: Power = 500hp = 372000w
Mass = 1750kg
Distance = 410m
Time = 11.5s

2. Relevant equations
Work = Force*Distance
Power = Work/Time

3. The attempt at a solution
Power = Force*Distance
372000w = (Force*410m)/11.5s
372000w = (410x N/m)/11.5s
Multiply each side by 11.5s
427800w = 410x N/m
Now i eliminated time, work should be 427800N, but that does not look correct. Need help and explanation here.

P.s. There will be no problems solving question b since if i get the work done I can divide it by the time.

See

2. Jun 6, 2015

### jbriggs444

I had some misgivings about this problem at first and had to do some sanity-checking on it. It seemed possible that you had been given either too much or too little information. It turns out that you have been extra information, but the numbers are consistent. [The approach that I took for the sanity check was to determine energy as a function of time, convert to velocity as a function of time, integrate to get distance as a function of time and check that the indicated distance was consistent with the indicated elapsed time].

The problem with this is that power is not equal to force times distance. Work is equal to force times distance. But that's not quite right either. Work is equal to force times distance for a constant force. We do not have a constant force here. Instead we have constant power. In order for power to remain constant as velocity increases, force must decrease (as the Mustang goes faster it has to run through higher and higher gears).

What is the definition of power?

If you could determine the energy of the Mustang after 11.5 seconds had elapsed, would this allow you to compute its final velocity?

3. Jun 6, 2015

### MrAnchovy

Yes jbriggs444, at first glance I thought this was a conservation of energy too, but at second glance:
Can you use these two equations to get a single equation for force related only to power, distance and time (so you don't need to calculate work)?

4. Jun 6, 2015

### jbriggs444

As I already wrote, work is not equal to force times distance in this case. Toss that formula straight into the trash can. Not applicable -- unless you are interested in doing some calculus.

5. Jun 6, 2015

### Seejianshin

Oh boy so sorry jbriggs444 that problem was a typo! I always knew my teacher warned me about being careful.

What i sould had wrote is power = force*distance/time.

And about the constant force i think the question only wanted me to use average force(i think).

6. Jun 6, 2015

### MrAnchovy

jbriggs444 I assumed that the question was asking for the force averaged over distance in which case total work done IS equal to average force times distance - surely at this level it is appropriate to assume that (although you do of course need calculus to prove it)? Isn't the alternative - to calculate force averaged over time - too complicated for this level?

7. Jun 6, 2015

### jbriggs444

I assumed the opposite. That the problem is asking for force averaged over time. It does not matter much. The approach is the same in either case and is equally simple.

8. Jun 6, 2015

### MrAnchovy

Ah no, jbriggs444 you are right, using the hint in the question to first calculate work done enables the average force over time to be calculated simply. Ignore my method.

9. Jun 6, 2015

### jbriggs444

No. The question does not want you to use average force. The problem wants you to compute average force. The tricky part is that the problem gives you more information than you need. Start with power and time. Ignore distance. See where that leads to start with.

10. Jun 6, 2015

### haruspex

If it doesn't specify 'averaged over distance' and the force is not constant, averaging over distance will give the wrong answer. See section 3 of https://www.physicsforums.com/insights/frequently-made-errors-mechanics-forces/.

11. Jun 7, 2015

### Seejianshin

Thanks again jbriggs444 i read the question again and i realized that i over did.
Since power = work/time
so this is what i did.
372000w= Work/11.5s
(multiply both sides by 11.5s)
Work = 4278000J
Thanks!
Just correct me if im wrong, question b) should be:
Work = Force*Distance
4278000J = Force*410m
(divide both sides by 410)
Force should be 4278000/410 = 10434.1463415N

12. Jun 7, 2015

### haruspex

Strange... assuming no losses, I get distance as $\sqrt{\frac{8Pt^3}{9m}}$, giving 536m. I can't see where I'm going wrong.

13. Jun 7, 2015

### haruspex

That way of working out average force is only valid if either "average force over distance" is specified or the force is constant. Neither is the case here. The correct way is to find the average acceleration (increase in velocity divided by elapsed time) and multiply by the mass. See the link In post #10.
In the present problem, this may be awkward. We can only compute the final velocity from the given information if we can somehow know that there are no losses, i.e. that all the energy delivered by the engine has turned into KE. I gather from jbriggs' posts that he concludes that is the case since otherwise the given distance could not have been covered, but as you can see from my posts I am not currently able to make the same deduction. If my calculation is correct, there are losses and there is not enough information to deduce an average force.

14. Jun 7, 2015

### jbriggs444

I think I buggered the sanity check. Your number and mine were off by a factor of almost exactly 0.75. I re-did the derivation and agree with your formula.

15. Jun 7, 2015

### insightful

I get the distance as sqrt(0.5Pt^3/m) = 403. (<<<DELETE)

EDIT: Found my mistake (assumed constant v...red face).

Last edited: Jun 7, 2015
16. Jun 7, 2015

### haruspex

Rather than divulge the solution to the OP, I've sent you my algebra via a private conversation.