Determining the position from the velocity

[zakare]

Homework Statement

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

Homework Equations

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

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

 

[arildno]

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

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

 

[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

 

[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?

 

[tiny-tim]

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

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.

 

[Multicol]

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

edit: first post and I am late :(

 

[tiny-tim]

Welcome to PF!

Hi Multicol ! Welcome to PF!

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

edit: first post and I am late :(

that's ok … spring was early this year! ​