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Spring pushing a block

  • #1
A spring of spring constant k = 340 Nm-1 is used to launch a 1.5-kg block along a horizontal surface by compressing the spring by a distance of 18 cm. If the coefficient of sliding friction between the block and the surface is 0.25, how far does the block slide?

Well i have:

W=1/2kx^2
W=55080J
W=U(potential energy) so i know the potential energy of the spring is 55080
Energy is conserved..so after the spring has decompressed all the energy is in the block..which is now kinetic
K=1/2mv^2
55080=1/2(1.5)v^2
so Vi=406.5m/s ?? that seems pretty darn high, this is just the initial velocity right?
so Vf=0
friction = 3.7N
F=ma
Vf^2=vi^2+2ax

I assume i'm using these equations..but am not sure how i arrise to get accel or if i'm on the right track..
 

Answers and Replies

  • #2
454
0
The spring constant is given in Nm-1 so you have to convert the length of the spring to
meters. you also multiplied by the mass (1.5kg) when you should have divided when calculating Vi.
Calculating the speed of the block is really unnecessary, because you can do it all with potential energy. All of the potential energy of the spring will go to friction. Since the friction force is constant and in the opposite direction of the velocity. The energy dissipated by friction is Force * distance.
(using a rounded off value of 3.7N in your calculations could make the final answer wrong)
 
  • #3
The spring constant is given in Nm-1 so you have to convert the length of the spring to
meters. you also multiplied by the mass (1.5kg) when you should have divided when calculating Vi.
Calculating the speed of the block is really unnecessary, because you can do it all with potential energy. All of the potential energy of the spring will go to friction. Since the friction force is constant and in the opposite direction of the velocity. The energy dissipated by friction is Force * distance.
(using a rounded off value of 3.7N in your calculations could make the final answer wrong)

well..friction is 3.675N
So, i know the potential energy of the spring is 55080J, how would i set up an equation to figure out the distance the block would travel?
 
  • #4
Doc Al
Mentor
44,877
1,129
Hint: All the initial energy ends up dissipated by friction. What work is done by the friction force?
 
  • #5
Hint: All the initial energy ends up dissipated by friction. What work is done by the friction force?
The work of friction (Wf) is:
W=Fx
so Wf=3.675x
 
  • #6
Doc Al
Mentor
44,877
1,129
The work of friction (Wf) is:
W=Fx
so Wf=3.675x
Good. Keep going.
 
  • #7
the work of the spring (potential energy) (using the work-energy theorem) is 55080...
so 55080=3.675x
x=14987.75?
that can't be right..i must be missing a step?
 
  • #8
Doc Al
Mentor
44,877
1,129
the work of the spring (potential energy) (using the work-energy theorem) is 55080...
Redo this calculation, paying attention to units. (Convert cm to m.)
so 55080=3.675x
x=14987.75?
that can't be right..i must be missing a step?
Once you correct your value for spring PE you should be fine. I don't see any missing steps. (Calculating KE at intermediate points is unecessary.)
 
  • #9
Redo this calculation, paying attention to units. (Convert cm to m.)

Once you correct your value for spring PE you should be fine. I don't see any missing steps. (Calculating KE at intermediate points is unecessary.)


BLAGH!!
stupid mistake on my part...
so work now is equal to 5.5J
so 5.5=3.675x
x=1.49m

answer to the problem..
 

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