What is the Conservation of Energy in a Car Braking on an Icy Road?

[dralion87]

a 1500kg car slams on the brakes on an icy road with a coefficient of kinetic friction of .05. the car has initial velocity before applying the brakes.find the distance requires for the car to stop completely. use the principles of conservation of energy.

 

[konthelion]

You need to show your work.
Start with the definitions. The mechanical energy lost is $$f \Delta s=\Delta E_{therm}$$, where f is the magnitude of the friction force(what is the definition of kinetic friction?), and $$\Delta s$$ is the distance that you are trying to find.

By the work-energy theorem, $$W_{ext}=\Delta E_{mech}+\Delta E_{therm}$$. Now, solve for $$\Delta s$$

 

[dralion87]

yea i need to show work... but what is delta mech and delta therm i don't know how to set up the problem..

 

[konthelion]

I just gave you the definition of $$\Delta E_{therm{$$; $$\Delta E_{mech}$$ is just the mechanical energy i.e. the initial kinetic energy. Open up your textbook and look at the definitions.

 

[dralion87]

im lost really i just don't get it..would you kbe kind enough and plug in nummbers

 

[konthelion]

Absolutely not. You haven't done anything. As for the answer, it's very easy to see you just have to do it yourself. What is the definition of kinetic energy? What is f?

 

[dralion87]

kinetic energy is (1/2) M V^2 and F= friction

 

[konthelion]

No, I meant force of friction. What is the definition of force of friction? After you've done that, then you can plug it in the equation that I told you:

$$W_{ext}=\Delta E_{mech}+\Delta E_{therm}$$
You know,
(1) $$f \Delta s=\Delta E_{therm}$$
(2) $$W_{ext}=\Delta U + \Delta K + f\Delta s$$
(2) As you've just said, $$\Delta K =\frac{1}{2}mv^2$$

Now, ask yourself, what must $$W_{ext}$$ be? Then write the definition of f. Then find s.

 

[dralion87]

this is wat i have

(1/2)(1500)v^2+.05 delta s =.05 delta s

 

[konthelion]

No.You are given the coefficient of friction,, which is denoted as "mu" $$\mu$$. What is the friction force? $$f=\mu*blank$$ What is blank? It's a definition; look it up in your book.

Where did you get the terms on the right-hand side? What is $$W_{ext}$$? Hint: There are no conservative forces and no external forces acting on the system, then W=?

 

[dralion87]

its equal to N which is mass times gravity ...i been on this problem for almost 3 3hours long and i got like 5 more to go and its due tomorrow please please please please

 

[konthelion]

Ok, now plug it back into the equation, we get

$$W=\frac{1}{2}mv^2+\mu mg \Delta s$$

You're almost there. What must W be? Then you can solve for $$\Delta s$$

 

[dralion87]

w must be the same as (1/2)mv^2 + "mu"mg

 

[konthelion]

No. If there are no conservative forces doing work and no external forces acting on the system, then W=?

 

[dralion87]

it should equal 0

 

[dralion87]

it should equal 0

 

[konthelion]

Yes! Now, solve for the distance, $$\Delta s$$

 

[dralion87]

i got 1.02m i don't think it my answer makes sense

 

[konthelion]

What do you mean it doesn't make sense? Show your work.

 

[dralion87]

alright this is wat i put in... (1/2)(1500)(cos180)^2 divided by (.05)(1500)(9.8)

 

[konthelion]

That makes no sense whatsover, plug W=0 back into the equation.

$$W=0=\frac{1}{2}mv^2+\mu mg \Delta s$$
Solve for $$\Delta s$$ I fail to see the issue here.

 

[dralion87]

is v^2 equal to cos 180... or what do i pug in for velocity since its not given

 

[konthelion]

Obviously, you won't get a numerical value and $$v^2 \neq cos(180)$$(where are you getting this from? this is nonsense). You are not given the numerical value of the initial velocity, so of course, you won't get a numerical value for the distance.

Look, use your ALGBEBRA skills to solve for $$\Delta s$$. Anything else, you're wasting your time.

You are close, you have the $$\Delta x$$ as some constant * $$V_{i}^2$$

 

[dralion87]

this is wat i got then (-750v^2) / (735) =delta s

 

[konthelion]

Solve it algebraically, then plug in numbers. You will see that mass does not matter.

$$\Delta s =\frac{1}{2\mu g} *v_{i}^2$$

 

[dralion87]

ok this is what i get

(1/2)MV^2 / "mu"MG = delta s

 

[konthelion]

Right, so the "M"s cancel.

 

[dralion87]

so i get (1/2)V^2 / "mu" G = delta s