# Homework Help: Spring compression homework problem

1. Oct 28, 2013

### nkentros7

1. The problem statement, all variables and given/known data
A 165g plastic block is set up against a spring. The block rests on a smooth (I'm assuming this implies that there is no friction involved) horizontal surface. The block is pushed into the spring, compressing it a distance of 15.0 cm and released. The spring constant is k= 677 N/m
(a) What is the speed of the block when it leaves the spring?
(b) If the table is 0.875m in height, what horizontal distance does the block travel after it leaves the table edge?

2. Relevant equations
After a bit of research, I found Hooke's Law: F=kx (where k is the constant provided in the problem statement)
F=ma
Perhaps one of the kinematics equations; Vf^2 = Vi^2 + 2a(delta X)

3. The attempt at a solution
The first thing I did was use the value given for k and the compression distance of 15 cm (converted to .15m for unit accuracy) to solve for the force, which came out to be 101.55N. With this, I can solve for acceleration, as we were given the mass. I get stuck here: how can I solve for initial velocity with only acceleration and force? I thought about using the distance that the string was compressed as a value for delta X, but that still leaves me missing data needed to solve for the initial velocity in one of the kinematics equations. I haven't tried assuming the information given in part B as part of A yet, but maybe there was a mistake on my teacher's part (including information in part B that is necessary for part A)

Last edited: Oct 28, 2013
2. Oct 28, 2013

### nil1996

The block is not attached to the spring.So it will leave the spring after it has traveled the 15cm.So now you can calculate the velocity at this moment.

Welcome to forum

3. Oct 28, 2013

### nkentros7

Thank you for the welcome. What extra information does this give me to solve the problem, though? That situation still leaves me with no time, no initial/final velocity, and in this case, no distance.

4. Oct 28, 2013

### nil1996

See the force due to the spring will cause acceleration in the block.This force will cause acceleration for the 15cm.Because the block will leave the spring after it has traveled the 15cm.Now you have to first calculate accelearation.Then use the formula Vf^2 = Vi^2 + 2a(delta X).Take initial velocity zero.

5. Oct 28, 2013

### nkentros7

Initial velocity is not zero. I am solving for the initial velocity after the object is released from the spring.
Anyways, I think I figured it out; had another look at an equation sheet. Thanks for your help.

Just to check my work, I used the equation PEspring = (1/2)kx^2. With the total distance, I can now solve for initial velocity because I can assume now that final velocity will be 0.

EDIT: Actually, I am using PEelastic = 1/2 k x^2 using the spring displacement of .15m.
Still have questions: how can I relate this elastic potential to any kinematic values? Would I use the principle that the change in potential energy is equal to work, and solve for total distance using this information?

Last edited: Oct 28, 2013
6. Oct 28, 2013

### nil1996

Initial velocity is zero as the block is at rest for some time before it starts running on the table.
How final velocity will be zero?What do you mean by final velocity?

7. Oct 28, 2013

### nkentros7

The block is under the effects of a force as it is attached to a string that is being compressed. When this force is released, the block will be pushed by the string at a certain initial velocity.

If I had the total distance that the block would travel, I could assume that final velocity = 0, because if the block is not increasing in distance, its velocity must be 0. However, I realized I couldn't find total distance.

8. Oct 28, 2013

### nil1996

work enegy theorem works quite well

Last edited: Oct 28, 2013
9. Oct 28, 2013

### nkentros7

Got it! Saw that total energy = elastic potential + kinetic energy. Thus, when fully compressed, all of the systems energy is equal to 1/2 k x^2. In this case, this is equal to 7.62. When the system has been released and the spring has stretched back to its normal point (where the displacement of the string = 0), the object's kinetic energy is at its maximum and its elastic potential = 0. It follows that our total energy (7.62 J ) = 1/2 (m)(vi^2), with which we can solve for Vi = 9.6 m/s.

10. Oct 28, 2013

### nil1996

That is the final velocity not the initial man

11. Oct 28, 2013

### nkentros7

Sorry, I disagree. The initial velocity is not 0 for any fraction of a second. As soon as the force holding the spring is released, the object is immediately propelled at a certain velocity. The object begins from rest, but is under a elastic force that will push it as soon as the other force holding it in equilibrium is removed. If a unequal force is acting on an object, there is no point where the object will not be accelerating.

12. Oct 28, 2013

### nil1996

As you wish.Go on.....

13. Oct 28, 2013

### nkentros7

Depends on how you look at it. The problem is asking for the velocity as the block leaves the spring, which is exactly the point where there is no more elastic potential and the block is at maximum kinetic energy. It is the final velocity for its path on the spring, and the initial velocity for its path afterwards.

14. Oct 28, 2013

### nkentros7

Sorry, I know you were trying to help. I'm pretty confident in my reasoning here, though. Thanks for your efforts =)

15. Oct 29, 2013

### haruspex

I confirm that answer. However, I do agree with nil1996 that calling this the initial velocity is rather confusing, given that it does not slow down as it traverses the table.