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Spring force speed of block

  • #1
A 2.00-kg block is pushed against a spring with negligible mass and force constant k = 400 N/m, compressing it 0.220 m. When the block is released, it moves along a frictionless, horizontal surface and then up a frictionless incline with slope of 37 degrees.
a) What is the speed of the block as it slides along the horizontal surface after having left the spring?
b)How far does the block travel up the incline before starting to slide back down?

on part a i tried using the equation final velocity^2= initial velocity^2 +2*acceleration*distance, but i am getting no where. I really can't do b without finding the speed.
 

Answers and Replies

  • #2
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Draw a free body diagram. Then, use conservation of mechanical energy.
 
  • #3
Doc Al
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That kinematic equation only works for cases of constant acceleration, but that doesn't apply here since the spring force (and thus corresponding acceleration of the block) is not constant.

Luckily, something is conserved that makes finding the speed of the block after it leaves the spring easy to calculate. What is conserved?
 
  • #4
the mass and energy?
 
  • #5
Doc Al
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the mass and energy?
Yes. Energy conservation is what you want to use here. What's the energy stored in a compressed spring?
 
  • #6
elastic potential energy which is .5kx^2 ?
 
  • #7
Doc Al
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elastic potential energy which is .5kx^2 ?
Exactly. That's the total mechanical energy of the system.
 
  • #8
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couldnt we just break the problem up into 2 sections, one where the block is compressed and then released and the other when the block is climbing the inclined plane? maybe you guys were already thinking this and im just being redundant.

its been a semester since my last physics class and i thought id just do this for fun so when i solved for height, is that the height from the ground, or is that the length it climbs on the angled part of the inclined plane?

for the first half i just did (elastic potential)=(kinetic energy final)

and on the second i did (kinetic energy initial [the initial from the first half])=(gravitational potential final)
 
  • #9
238
0
couldnt we just break the problem up into 2 sections, one where the block is compressed and then released and the other when the block is climbing the inclined plane? maybe you guys were already thinking this and im just being redundant.

its been a semester since my last physics class and i thought id just do this for fun so when i solved for height, is that the height from the ground, or is that the length it climbs on the angled part of the inclined plane?

for the first half i just did (elastic potential)=(kinetic energy final)

and on the second i did (kinetic energy initial [the initial from the first half])=(gravitational potential final)
For part (b), how far it goes up would be the distance that the block travel before momentarily coming to rest, let's call s.

(1) [tex]E_{mech,i}=U_{spring,i}+U_{g,i}+K_{i}[/tex]
(2) [tex]E_{mech,f}=U_{spring,f}+U_{g,f}+K_{f}[/tex]


Four terms in equations (1) and (2) will be equal to zero(one of which is the initial kinetic energy [tex]K_{i}[/tex]) since the only forces that do work are the force exerted by the spring and the force of gravity. Then you are able to solve for h, the distance in height. To solve for the distance s(in terms of h), you can use trig since you are giving the angle.
 
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