EK Physics Lecture 2 #44 (Lecture Question)

[PhaZed]

---

Members don't see this ad.

Hello,

This might seem like a dumb question, but I am having problems figuring out how the book derived the following answer to the question below.

#44 "On a particular stretch of wet pavement, the kinetic coefficient of friction u for a particular car with mass m is 0.08. If the car is moving at a velocity v, and suddenly locks its wheels and slides to a stop, which of the following expressions gives the distance that it will slide?"

I know we will be using the following equation: Fk = UkFn

The normal force Fn is equal to the w of the car (mg).
Fk is equal to ma, so the acceleration is ug

If we plug that acceleration into the formula, V^2 = Vo^2 + 2ax the book states that the answer is (v^2 / 2gu).

Can someone please explain how they plugged it in the formula and derived the answer?

Thanks

 

[Rabolisk]

0 = v0^2 - 2μgx
2μgx = v0^2
x = v0^2/2μg

Basic algebra?

 

[attixx]

You pretty much have the answer.

V^2 = Vo^2 + 2ax

V is zero because is comes to rest, so you can rearrange for essentially V^2 = 2ax (the new V is Vo from the equation that you moved to the other side, but acceleration is negative so negatives cancel)

x=V^2/2a --- and you solved for a already so a=ug. plug in and...

x=V^2/2ug

 

[ilovemcat]

I would of solved in a different way:

The car is driving at some velocity. When it locks it's wheels and begins to slide, it's experiencing friction (opposing it's motion). This sliding friction is in turn slowing the car down now. The car approaches a stop. Therefore Vf = 0 m/s. It's initial velocity is just Vi, which we don't know since they didn't specify.

Using the Work-Energy Theorem:
Fd = 1/2mv^2 (where F is friction)(v is initial velocity)
F = (uk)(N) = (uk)(mg)

Plugging that in and solving for d yields:
d = 1/2v^2/(uk)(N); N = mg so...
d = 1/2v^2/(uk)(mg) or v^2/2(uk)(mg)

--------------

If you instead used: Vf^2 = Vi^2 + 2ad (Vf = 0 m/s because the car will come to a stop).
Re-arranging yields: -2ad = Vi^2
Re-arranging again: d = Vi^2/(-2a)

Solving for a:
Fnet = -Friction (because it's opposing velocity, we add a minus sign)
Fnet = -ukN = -ukmg
ma = -ukmg
a = -ukg

Finally, substituting yields:
d=Vi^2/2(ukg)

 

[ilovemcat]

When I was writing, nobody replied. Looks like I'm a little late, lol. 😛

 

[PhaZed]

Haha... thank you. I feel like an idiot.