Work and Energy with 2 variables

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Homework Help Overview

The discussion revolves around a physics problem involving work and energy, specifically related to a car's skid marks and the forces acting on it during braking. The problem includes parameters such as the distance of the skid mark, the coefficient of friction, and gravitational acceleration.

Discussion Character

  • Exploratory, Assumption checking, Conceptual clarification

Approaches and Questions Raised

  • Participants discuss the relationship between kinetic energy and work done by friction, with attempts to derive the car's initial speed from the given data. Questions arise regarding the necessity of the car's mass and the interpretation of work and energy equations.

Discussion Status

Some participants are exploring the equations related to kinetic energy and work, while others are questioning the validity of the derived speed based on the skid distance. There is acknowledgment of potential errors in reasoning and the need for further clarification on specific terms used in the equations.

Contextual Notes

Participants note a possible lack of information or misinterpretation of the problem setup, particularly regarding the expected speed of the vehicle and its implications on the skid distance.

zenophysics
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Homework Statement


I feel as though I am missing information but I have been assured I'm not can anyone help me out?

At an accident scene on a level road, investigators measure a car's skid mark to be d=88m long. The accident occurred on a rainy day, and the coefficient of friction was estimated to be Mu=0.34. Use this data to determine the speed of the car, v when the driver slammed on (and locked) the breaks. Take g=10m/s^2.


Homework Equations


What is the speed of the car?
F=ma Ft=m(delta)v


The Attempt at a Solution


E1=E2+|W*Ffr|
mv^2/2=mv^2/2+|W*Ffr|
 
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zenophysics said:

The Attempt at a Solution


E1=E2+|W*Ffr|
mv^2/2=mv^2/2+|W*Ffr|

Initially the car has KE mv2/2

when the car has gone a distance of 88 m the velocity is zero.

So applying your second formula you will get

mv2/2 = Work done by friction.

You should know how to get the frictional force (so you will see why you don't need the mass of the car).
 
ok so if mv^2/v = WFfr and Ffr is the same as Fn*mu = mg*mu in the opposite direction then mass cancels out and leaves me with v^2/2=g*mu and when I solve for V I'm getting v=2.89ms. yes?
 
You're on the right track, but I see one error in your attempt to solve the problem.

Namely the statement in bold here:
E1 = E1'+W*Ffr

As a quick review, we know that

W = Fd

and

W = KE

Thus,

KE = Fd

Well, heat caused by friction is a form of kinetic energy, so we can surmise the following:

Eh = Ffrd

Can you see how this can apply to your equation?

Please note, however, that this only accounts for the transfer of your original kinetic energy into heat energy. That isn't a problem in introductory physics, but sound and other forms of energy may be important in later years.
 
zenophysics said:
ok so if mv^2/v = WFfr and Ffr is the same as Fn*mu = mg*mu in the opposite direction then mass cancels out and leaves me with v^2/2=g*mu and when I solve for V I'm getting v=2.89ms. yes?

2.89m/s is equivalent to 6.46 mi/h (10.4 km/h), it seems unlikely that a vehicle would skid 88 meters with that kind of velocity.

Logic would conclude that your answer is missing a factor.

Where did you get the term " W*Ffr " and what does it define?
 

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