How far does the car slide?

  • Thread starter kylejustin
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I am stuck on the last problem for my homework. It seems to me like the problem doesn't have enough information to complete. I'd really appreciate any tips.

A car traveling along a level road as a speed V slams on the brakes and skids to a stop. If the force of friction on the car is half the car’s weight, how far does the car slide? (Hint: Use work-energy theorem and solve for d)

OK, so I know the following:

W=Fd

KE=1/2mv^2

Work=Change in KE

To use the work energy theorem, I need to know the KE. The find the KE, I meed to know either the mass of velocity. The problem tells me that the force of friction is 1/2 the car's weight, but that still doesn't give me any numbers to work with. This is where I'm stuck.
 

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LeonhardEuler

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It is common to want to work with numbers but here, as is often the case, it is better to work the problem out with symbols and plug the numbers in in the last step. Keep the car's mass as "m". By the time the problem is over (which is not a long time) the "m" will cancel out and you will have your answer.
 
W = Fd
KE = 1/2mv^2

Work = Change in KE

The change in KE is asking for the derivative of KE, because change over time is another application of the derivative. The derivative d/dv for (1/2)mv^2 is Change in KE = mv

Therefore you can plug the change in KE into the final equation

mv = Fd and solve for d.
 
First, thanks for the tips, but I just am not getting it. Is my final answer going to be a number?
 

LeonhardEuler

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It won't be a number, it will be an expression in terms of the given speed V and the acceleration due to gravity, g.

Do you know how to express the friction force in terms of m and g? Once you do its sraight foward to set the work equal to the change in kinetic energy and solve for d.
willworkforfood said:
The change in KE is asking for the derivative of KE, because change over time is another application of the derivative. The derivative d/dv for (1/2)mv^2 is Change in KE = mv
The change in KE is not the derivative of the KE. The rate of change of KE with respect to some other variable would be the derivative, but not the total change.
 

HallsofIvy

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Since the car has an initial speed of v, its initial kinetic energy is (1/2)mv2. When it stops (v= 0) its kinetic energy is (1/2)m02= 0. Okay, the work done (by the friction) must be (1/2)mv2.

You are told that the friction force is "half the cars weight" which is (1/2)mg. (g= 9.81 m/s2.) You find the work done by integrating force time distance which with constant force is "force times distance" or (1/2)mgd. If (1/2)mgd= (1/2)mv2, what is d?
 

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