# Velocity related to force of friction

1. Jan 12, 2005

### UrbanXrisis

A 5kg object slides 100m down a frictionless inclined plane dropping 45m. It then slides along a horrizontal surface with a coefficiet of kinetic friction of .75 intil it stops. How long does it take to stop after it leaves the inclined plane?

I'm not sure how to get the time. Here's what I found:

mgh=.5mv^2
10m/s/s*45m=.5 v^2
v=30 m/s

So the object is going 30m/s when it hits the friction floor. I'm not sure what to do next.

2. Jan 12, 2005

### derekmohammed

Hi there,

This is a conservation of Energy question.

The most important thing to do is to set up the question right.

We now that the Ek = Ethermal (Ethermal is the energy lost to the fristional force)

Ek= squrt(2gh) becuase the Ek will be resulting from the drop on the inclined plane and the 45m drop.

Ethermal= mgd0.75

squrt(2gh)=mgd0.75
Solve for distance then if you the velocity just use V=d/t to find the time!

3. Jan 12, 2005

### Curious3141

To determine the kinetic energy of the object as it leaves the inclined plane, use conservation of energy.

Now you have the kinetic energy given by $E_k = \frac{1}{2}mu^2$. Get the value of $u$ and then plug it into $v = u + at$ where $v = 0$ and $a = - \frac{F_{fr}}{m} = - \mu g$ and solve for $t$.

EDIT : I misread the question, they asked for "how long" (time) not distance.

Last edited: Jan 12, 2005
4. Jan 12, 2005

### dextercioby

I'm sorry,but your last equation doesn't make any sense.You equate velocity to work done by friction force...

Daniel.

5. Jan 12, 2005

### Curious3141

You're talking about my post or derek's ?

6. Jan 12, 2005

### dextercioby

You hadn't written your reply,by the time i got into writing mine,so that's why i didn't quote the mistake.There it is:
Daniel.

7. Jan 12, 2005

### derekmohammed

Opps. :rofl:

I just finished a two hour lecture on aristotle :zzz: when I wrote this post.

I see what I did I am :uhh: sooooo sorry. What I really meant was that EK = Ep and that you could find Velocity through that Equation.... Sorry. the final Equation should be:

mgh = mgd0.75

Where h is the total distance the box falls!

SORRY

Last edited: Jan 12, 2005