Work and Energy

  • Thread starter HumorMe81
  • Start date
  • #1
17
0
A light horizontal spring has a spring constant
of 108 N/m. A 3.05 kg block is pressed against
one end of the spring, compressing the spring
0.111 m. After the block is released, the block
moves 0.204 m to the right before coming to
rest.
The acceleration of gravity is 9.81 m/s2 .
What is the coefficient of kinetic friction
between the horizontal surface and the block?

This is how I solved it, although I don't think it's correct.

When the spring is compressed, it has some potential energy and kinetic energy at this point is 0:
spring compressed:
u=1/2kx^2
u=1/2(108N/m)(.111m)^2
U=.67J

Ke=1/2mv^2
I don't know the acceleration in the horizontal acceleration so I can't figure out the velocity.
The problem also says the block moves .204m before coming to stop.
When the block comes to stop, all of the potential energy has been converted into kinetic energy and some has been lost to friction.
W' (work done by friction)
W'=Uf+Kef=Fk*d
=0+Uinitial, when compressed =Fk*d
=0+ .67J=mu k*N*d (N=mg for obj. on flat surface)
.67J=mu k (29.89N)(.204m)
muK= .11

any suggestions??
Thanks
 

Answers and Replies

  • #2
1,097
3
As you compress this block against your spring you aren't going to have any initial kinetic energy. Why? Because the block hasn't started moving yet.

See if you can come up with a rough formula to describe the conservation of energy throughout this system.

HINT: Use [tex] KEnergy, PEnergy[/tex] and [tex] \Delta Eth [/tex]
 
  • #3
17
0
As you compress this block against your spring you aren't going to have any initial kinetic energy. Why? Because the block hasn't started moving yet.

See if you can come up with a rough formula to describe the conservation of energy throughout this system.

HINT: Use [tex] KEnergy, PEnergy[/tex] and [tex] \Delta Eth [/tex]


ok, if the energy is conserved then
ΔE=ΔU+ΔKe=constant or 0
 
  • #4
1,097
3
I'm not sure why you're using Delta's for all the other quantities of energy. Initially all your energy is stored in the spring. In the final state all you energy has been destroyed by friction. So,

[tex]Initial Potential Energy[/tex] = [tex]\Delta Thermal Energy[/tex]

Can you define both types of energy?
 
  • #5
17
0
I'm not sure why you're using Delta's for all the other quantities of energy. Initially all your energy is stored in the spring. In the final state all you energy has been destroyed by friction. So,

[tex]Initial Potential Energy[/tex] = [tex]\Delta Thermal Energy[/tex]

Can you define both types of energy?

I used deltas to define the initial and final energies of the entire syste.
Ei = Ui + KEi
= 1/2kx^2 + 0
So, the initial energy = 1/2kx^2 only b/c the block has not been released yet.
Finally,
Ef = (Uf + KEf) - work done by friction
This is how it makes sense to me.
my mcat book, however, says to use the formula:
W' = U + Ke = Fk*d
 
  • #6
1,097
3
In your final state you have no kinetic or spring energy so its safe to say that,

[tex]Uf=0[/tex] and [tex]Kf=0[/tex].

So,

[tex]\frac{1}{2}kx^{2} = \Delta Eth[/tex]

Simply define [tex]\Delta Eth[/tex]
 
  • #7
17
0
In your final state you have no kinetic or spring energy so its safe to say that,

[tex]Uf=0[/tex] and [tex]Kf=0[/tex].

So,

[tex]\frac{1}{2}kx^{2} = \Delta Eth[/tex]

Simply define [tex]\Delta Eth[/tex]

Ok, got it! Thank you so much!
W'(work done by friction) = delta E = Fk*d
 
  • #8
1,097
3
No problem, I'm assuming you take things from here.
 

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