Determining the position from the velocity

  • Thread starter zakare
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  • #1
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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
 

Answers and Replies

  • #2
arildno
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Was the force of friction given as 1N, or is that your assumption?

Insofar as it was given, remember the work-energy formula.
 
  • #3
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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
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I seriously doubt that you need all the info (but maybe I'm doing something wrong):

[tex] \frac{1}{2}mv^2 =mgh [/tex]

solving for v:

[tex] v = \sqrt{2gh} [/tex]

Inserting data I get 5 cm lower w.r.t initial height?
 
  • #5
tiny-tim
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… but I don't know how to add in the energie given by the spring and the energy lost from friction.

Hi zakare! :smile:

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. :smile:
 
  • #6
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It will also reach 1 m/s when its a bit in the spring

edit: first post and im late :(
 
  • #7
tiny-tim
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Welcome to PF!

Hi Multicol ! Welcome to PF! :smile:
It will also reach 1 m/s when its a bit in the spring

edit: first post and im late :(

:smile: that's ok … spring was early this year! :smile:
 

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