# Determining the position from the velocity

1. Jun 1, 2008

### zakare

1. The problem statement, all variables and given/known data

An object (masse m) is placed on a slope with an angle a at a height of h above the ground. There is no friction on the inclined plane. The inclined plane leads to a horizontal plane 2 meters long that ends with a spring with a constant of k. There is friction on the horizontal plane (but not the inclined plane). Where is the object when its velocity is equal to 1m/s?

m=1kg
h=0.5m
k=1000N/m
angle a= 30 degrees
F(friction)=1N

2. Relevant equations

E(kin)=0.5*mv²
W = E(cin final) - E(cin initial)

3. The attempt at a solution

W = 0.5*mv² - 0.5*m(0)²
=0.5*mv² = F*d
d=(mv²)/(2F)
=.5m

but I don't know how to add in the energie given by the spring and the energy lost from friction.

Any help would be appreciated thanks

2. Jun 1, 2008

### arildno

Was the force of friction given as 1N, or is that your assumption?

Insofar as it was given, remember the work-energy formula.

3. Jun 1, 2008

### zakare

Yes, it's given. And sorry my attempt is a bit poor, I just need a bit of help starting off in the right direction. Thanks

4. Jun 1, 2008

### dirk_mec1

I seriously doubt that you need all the info (but maybe I'm doing something wrong):

$$\frac{1}{2}mv^2 =mgh$$

solving for v:

$$v = \sqrt{2gh}$$

Inserting data I get 5 cm lower w.r.t initial height?

5. Jun 1, 2008

### tiny-tim

Hi zakare!

The energy absorbed by the spring is 1/2*kx², where x is the decrease in length of the spring;

and the energy lost from friction equals the work done against friction.

6. Jun 1, 2008

### Multicol

It will also reach 1 m/s when its a bit in the spring

edit: first post and im late :(

7. Jun 1, 2008

### tiny-tim

Welcome to PF!

Hi Multicol ! Welcome to PF!
that's ok … spring was early this year!