How is Work Calculated for a Drag Race Scenario?

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Homework Help Overview

The discussion revolves around calculating work and average force in a drag racing scenario involving a 500hp Mustang and a 1750kg car over a 410m distance in 11.5 seconds. Participants explore the relationships between power, work, and force, questioning the assumptions and definitions involved in the problem.

Discussion Character

  • Exploratory, Conceptual clarification, Mathematical reasoning, Assumption checking

Approaches and Questions Raised

  • Participants discuss the relationship between power, work, and force, with some attempting to derive equations that relate these quantities without directly calculating work. Questions arise about the validity of using average force over distance versus time, and the implications of constant versus variable forces in the context of the problem.

Discussion Status

The discussion is active, with various interpretations being explored regarding how to approach the calculations. Some participants suggest that the problem provides more information than necessary, while others emphasize the importance of correctly identifying the type of average force required. Guidance has been offered on how to think about the relationships between the quantities involved, but no consensus has been reached.

Contextual Notes

Participants note potential issues with the information provided, including concerns about whether the problem assumes constant force or average force, and the implications of energy losses in the scenario. There is also mention of the need for calculus to accurately address the problem, indicating a level of complexity that may not be suitable for all participants.

Seejianshin
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Homework Statement


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

Homework Equations


Work = Force*Distance
Power = Work/Time

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.

Thanks in advance,
See
 
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Seejianshin said:

Homework Statement


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
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 Attempt at a Solution


Power = Force*Distance
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?
 
Yes jbriggs444, at first glance I thought this was a conservation of energy too, but at second glance:
Seejianshin said:

Homework Equations


Work = Force*Distance
Power = Work/Time
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)?
 
MrAnchovy said:
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)?
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.
 
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).
 
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?
 
MrAnchovy said:
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?
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.
 
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.
 
Seejianshin said:
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).
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.
 
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  • #10
MrAnchovy said:
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?
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
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 I am wrong, question b) should be:
Work = Force*Distance
4278000J = Force*410m
(divide both sides by 410)
Force should be 4278000/410 = 10434.1463415N
 
  • #12
jbriggs444 said:
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].
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
Seejianshin said:
Just correct me if I am wrong, question b) should be:
Work = Force*Distance
4278000J = Force*410m
(divide both sides by 410)
Force should be 4278000/410 = 10434.1463415N
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
haruspex said:
Strange... assuming no losses, I get distance as ##\sqrt{\frac{8Pt^3}{9m}}##, giving 536m. I can't see where I'm going wrong.
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
haruspex said:
Strange... assuming no losses, I get distance as 8Pt39m−−−−√\sqrt{\frac{8Pt^3}{9m}}, giving 536m. I can't see where I'm going wrong.
I get the distance as sqrt(0.5Pt^3/m) = 403. (<<<DELETE)

EDIT: Found my mistake (assumed constant v...red face).
 
Last edited:
  • #16
insightful said:
I get the distance as sqrt(0.5Pt^3/m) = 403.
Rather than divulge the solution to the OP, I've sent you my algebra via a private conversation.
 

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