How much work must be done to stop a 980 kg car traveling at 108 km/h?

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Homework Help Overview

The discussion revolves around calculating the work required to stop a car and determining the speed of a car based on skid marks and friction. The subject area includes concepts from mechanics, specifically kinetic energy and frictional forces.

Discussion Character

  • Exploratory, Conceptual clarification, Mathematical reasoning, Problem interpretation

Approaches and Questions Raised

  • The original poster attempts to calculate the work needed to stop a car using kinetic energy and expresses confusion about the correctness of their answer. They also seek help with a second problem involving skid marks and friction, questioning the relevance of mass in the calculations.
  • Some participants question the accuracy of the original poster's calculations and suggest writing out relevant equations to clarify the relationship between friction and acceleration.
  • Others explore the connection between the frictional force and potential energy, attempting to derive the speed from the given data.
  • One participant expresses uncertainty about their approach and calculations, indicating a need for further clarification.

Discussion Status

The discussion is ongoing, with participants providing various insights and equations related to the problems. Some guidance has been offered regarding the relationship between mass and friction, and there is a recognition of the need to incorporate distance into the calculations. Multiple interpretations of the problems are being explored, but no consensus has been reached on the correctness of the answers.

Contextual Notes

Participants are working under the constraints of homework rules, which may limit the information they can use or the methods they can apply. There is also a mention of uncertainty regarding the correctness of the teacher's guidance on the relevance of mass in the second problem.

TriumphDog1
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How much work must be done to stop a 980 kg car traveling at 108 km/h?
what i did was first convert the km/h to m/s:
[tex]108 * 1000 / 3600 = 30[/tex]
then plug the mass and velocity into the kinetic energy forumula [tex]K.E. = \frac{1}{2} mv^2[/tex]
[tex]\frac{1}{2}980(30^2) = 441000[/tex]
I have checked and rechecked it and i keep coming up with that answer but its wrong. Where am I going wrong or is there more to do?

Another problem i need help with is this:
At an accident scene on a level road, investigators measure a car's skid mark to be 88 m long. It was a rainy day and the coefficient of friction was estimated to be 0.46. Use these data to determine the speed of the car when the driver slammed on (and locked) the brakes.
I don't even know where to get started on this one especially without a mass, but my teacher says the mass doesn't matter.

Any help on these would be greatly appreciated.
 
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The first one seems right. Are you sure the answer is correct?
For the second one, write out the equation that relates friction force with acceleration, and you should see that the mass cancels out.
 
What's the correct answer for the first one?
As far as I can see - you've done it right.
 
yea - that's what i thought, but its not somehow, i think the teacher is wrong

im still not sure what to do on the second one,
Here's what i know:
[tex]F_{F} = \mu * F_{N}[/tex] or [tex]F_{F} = \mu * 9.8m[/tex]
and [tex]P.E. = .5mv^2[/tex]
and [tex]W = Fd[/tex] or [tex]W = mad[/tex]
those are the only formulas i know of that might help with this problem but I don't know what to do with them...
i tried setting the frictional force formala equal to the potential energy forumula and came up with:
[tex]\mu * 9.8 = .5v^2[/tex]
[tex].46 * 9.8 = .5v^2[/tex]
[tex]4.508 = .5v^2[/tex]
[tex]9.016 = v^2[/tex]
[tex]3.003 = v[/tex]
that was wrong, then i noticed i didnt even use the distance provided, but I don't know what else to do?
 
Last edited:
[tex] F_f = 4.5m[/tex]

so that means ...

[tex] F_{net} = ma - F_f[/tex]

so...

[tex] ma = 4.5m[/tex]

so...

[tex] a = 4.5[/tex]

using

[tex] v^2 = u^2 + 2ax[/tex]

v = 0 - because it stops, and u - is what you are trying to find...
take a = 4.5 and x = 88, so

[tex] u^2 = -2ax[/tex]

and i found that

[tex] u = 28.1[/tex]

I am not sure if this is right though
 
now - that i think about it - i think MY force equation is wrong..
 
thanks a lot i understand now... and that site is awesome
 

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