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Spring Potential Energy: Horizontal Spring Problem

  • #1

Homework Statement


A horizontal spring with a constant of 700N/m is compressed 2cm. A ball of mass 200g is placed in front of the spring. The spring is released. The ball slides along a horizontal track for 10cm while experiencing a frictional force of 0.75N. What is the velocity of the ball at the 10cm mark?

Homework Equations


[/B]
  • PEs= .5⋅Kx^2
  • .5⋅Kxi^2 + .5⋅m⋅Vi + mg⋅yi + Fnc⋅Δd= .5⋅Kxf^2 + mg⋅yf

3. The Attempt at a Solution

We just started learning about this in class the other day. Im not sure if the second equation is correct... but i know that the y-direction isnt included in this problem. K is the spring constant but I'm not sure what x is (Kx or Kxi)

If Kx is spring force do i multiply 700n/m and 0.75N? Im not really sure where to start here or which equation to use to find the velocity
 

Answers and Replies

  • #2
Chandra Prayaga
Science Advisor
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Homework Statement


A horizontal spring with a constant of 700N/m is compressed 2cm. A ball of mass 200g is placed in front of the spring. The spring is released. The ball slides along a horizontal track for 10cm while experiencing a frictional force of 0.75N. What is the velocity of the ball at the 10cm mark?

Homework Equations


[/B]
  • PEs= .5⋅Kx^2
  • .5⋅Kxi^2 + .5⋅m⋅Vi + mg⋅yi + Fnc⋅Δd= .5⋅Kxf^2 + mg⋅yf

The Attempt at a Solution


[/B]
We just started learning about this in class the other day. Im not sure if the second equation is correct... but i know that the y-direction isnt included in this problem. K is the spring constant but I'm not sure what x is (Kx or Kxi)

If Kx is spring force do i multiply 700n/m and 0.75N? Im not really sure where to start here or which equation to use to find the velocity
First, as you yoursekf pointed out, you must know what each symbol stands for and what each symbol means. It does not make much sense to just try to substitute some numbers into an equation.
Your second equation is certainly wrong. The second term in it should have a speed squared in it, not just the speed. The terms containing y are not relevant to this problem.
Have you done the work energy theorem ?
 
  • #3
RPinPA
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I always advise, "do the physics, then the equations, then the numbers".

First the physics. What's happening here? Energy is being given to the ball (by the spring), and energy is being taken away (by friction), and so whatever's left is the kinetic energy of the ball.

Now the equations. What is the expression for the energy the spring gives the ball?
What is the expression for the energy lost to friction?
What is the expression for the kinetic energy of the ball?

And now, finally the numbers. You know everything in that equation except the velocity of the ball at the end, the thing you're being asked for.
 
  • #4
haruspex
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You appear to have noted down some equations without a corresponding explanation of the variables.
The xi and xf are the initial and final displacements of the spring length from its relaxed length.

As @Chandra Prayaga points out, you have an error in the second equation. The velocity shoukd be squared. That may have happened hen you made your notes or when you wrote out your post.

@RPinPA gives good general advice but left out something important here. Sometimes equations only apply over certain parts of a process. In the present case, remember that the ball is not glued to the spring, so will lose contact when the spring has returned to its natural length. This means you have to solve the problem in two stages. First, consider the stage up to where contact is lost, then from there until the ball stops.
 
  • #5
CWatters
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  • #6
CWatters
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If Kx is spring force do i multiply 700n/m and 0.75N? Im not really sure where to start here or which equation to use to find the velocity
No.
I would start by working out the Potential Energy stored in the spring. Then think about where that energy will go. Write it down in English first. Then try and write equations, then rearrange them, then substitute numbers.
 
  • #7
RPinPA
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@RPinPA gives good general advice but left out something important here. Sometimes equations only apply over certain parts of a process. In the present case, remember that the ball is not glued to the spring, so will lose contact when the spring has returned to its natural length. This means you have to solve the problem in two stages. First, consider the stage up to where contact is lost, then from there until the ball stops.
You are absolutely correct and that complicates the equations somewhat. So for a fully correct answer you need to do it in two parts as you say.
1. The spring is still expanding and is in contact with the ball. At any distance d relative to the starting position, the spring has lost potential energy (write that expression) which has gone into friction (write that expression) and kinetic energy of the ball (write that expression).
This gives you a general equation you can solve for v anywhere along the way while expanding, in terms of the distance everything has moved.

When does it lose contact? When the spring's velocity is less than or equal to the ball's. So you need to find out at what d that happens.

2. You now know how much KE the ball has when it is moving on its own away from the spring. That energy is reduced by frictional work, leaving you the final KE at 10 cm.

Before doing any of that, you should ask yourself a really basic question. Does the ball EVER lose contact with the spring or does the spring go the whole 10 cm? You need to know what's happening as much as possible before deciding what equations describe it. The answer is "yes". The spring is compressed 2 cm, it will expand that 2 cm to its starting point and then at most another 2 cm beyond that. Less, because some of its energy was given to the ball. So the spring definitely won't go the entire 10 cm.
 

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