# How to calculate work done by spring?

1. Sep 18, 2009

### rustyphysics

Hi,

Was doing some testing for work, but a lot of my physics are gone! Hope someone can help...

Here goes:

A rigid mass is connected to a spring and the spring to a solid anchor on the ceiling. The mass has an initial velocity vi (due to free fall of the rigid mass). I have the force-time (F-t) chart and I am trying to determine the total amount of energy absorbed (work done is more appropriate here I suppose??) by the spring. How should I do this?

I was thinking of getting the area under the F-t chart and then multiply it by the average velocity. I assumed that the average velocity is (vi + vf)/2; vf is the final velocity = 0.

Initially this feels right, but after some thought, the approach looks suspicious. Can I assume that average velocity = (vi + vf)/2?

Any advice would be appreciated. Thanks!!

2. Sep 19, 2009

### granpa

the energy absorbed by the spring depends only on the distance it is stretched. the force alone tells you how far it has been stretched.

3. Sep 19, 2009

### SergioPL

Hi,

You say you have a spring hanging on the ceilling and it has an initial velocity when the spring is stretched ¿?.

The mass will oscilate interchanging energy between gravity and the spring.

The force done here is F = k·x -g·m

Where k is the spring constant, x is the distance that the spring has unstretched, g is the gravity constant and m is the mass.

So there is a relationship between the position and its second derivate that gives:

x = Csinωt

Where w = sqrt(k/m) and C is the maximum distance stretched

To know C, which is the maximum distance, you should convert the initial kinetic and potential energy in "spring energy" which is 0.5*k*x^2

So: 0.5*k*C^2 = C·m·g + 0.5·m·vi^2

Where vi is the "initial" kinetic energy.

Therefore v(t) = C·w·cos(wt)

and v_medium = C·w·2/pi

This means velocity in absolute value, of course :)

4. Sep 19, 2009

### rustyphysics

Thanks for the posts.

Unfortunately I am tasked to find out the amount of energy absorbed by the spring by using the force-time graph that I have. We are most concerned with the amount of energy absorbed when the spring first stretches until before it rebounds.

I found a similar post on this forum, but no answer was posted: https://www.physicsforums.com/showthread.php?t=204508

5. Sep 19, 2009

### Staff: Mentor

Are you expected to calculate the work done numerically from an empirical force-time graph?

The area under the F-t graph gives you the change in momentum. For a small enough interval, you can find the average speed and thus the distance traveled. From that you can find the work done in that interval.

6. Sep 20, 2009

### rustyphysics

Hi Doc Al,

Thanks for the post. Yes I'm suppose to determine the energy absorbed empirically. Your suggestion is useful. Let me give it a try... Cheers!

7. Sep 21, 2009

### DaTario

Hi

I suspect that mass will play an important role here, unless your F-t graph behaves well and vary to slowly. This context may call for an adiabatic elimination of the spring degree of fredom (or not if the force varies rapidly).
Energy of the system is kinetic plus potentials elastic and gravitational, and the kinetic will depend on the mass value and will depend on how large the force is and how fast force vary.

Best regards

DaTario

8. Sep 21, 2009

### sophiecentaur

Here's an alternative approach.
If your force time graph covers a number of oscillations then you can establish the spring constant, if the mass is known. (Ignoring the spring mass - but this could be included with more calculation.) this would tell you the spring PE at the bottom if you knew the amplitude of oscillation. The shape of the force time graph should tell you where the zero crossings of the sinwave are to give the amplitude.

9. Sep 21, 2009

### rustyphysics

Hi All,

Thanks for the interest. I came up with an excel spreadsheet to do some of the calculations based on Doc Al's suggestions. The calculations are described below

Raw data:
t=Time (ms)
dt = Time step (ms) [each time step is 1ms]
Fi=Force at time step i (kN)
M=mass (kg) [This is a fixed mass for each test]

Other variables:
ai = acceleration (m/s2) at time step i
vi = velocity (m/s) at time step i
si = displacement (m) at time step i
Wi = Work done (J) at time step i

For each time step:
ai = Fi x 1000 / M
vi = area under a - t graph = 0.5 (ai + ai-1) * 0.001 {1ms = 0.001s; convert ms to s}
si = area under v - t graph = 0.5 (vi + vi-1) * 0.001
Wi = Fi x si

The equations are pretty straightforward, but when I sum up the total work done, it seems to be too small. I would expect the total work done to be increased by a factor of 100. Some how I cannot spot my mistake, or is it due to some of the concerns that Da Tario mentioned?

sophiecentaur: Thanks for the suggestions. The graph does cover a few oscillations, but we are concerned that the Hooke's law may not apply as the "spring" is not exactly a spring.

Thanks.

10. Sep 22, 2009

### Staff: Mentor

Why are you multiplying by 1000?

Is the force in your F-t graph the total force (including gravity) or just the force of the spring? How did you measure this force?

11. Sep 22, 2009

### rustyphysics

Hi Doc Al

I was trying to convert the force in kN into N. Hence I multiplied F by 1000.

The Force in the F-t graph is a resultant force, therefore it includes gravity.

The measurement was based on a load cell that was connected to the "spring". The spring was connected to a mass that was dropped at certain height.

Thanks for the interest.

12. Sep 22, 2009

### Staff: Mentor

Do you have a diagram of the set up? I don't know much about load cells, but I would think that if it was attached to the spring it would measure tension in the spring (and thus spring force).

(I also wonder if this thread would be better off in Engineering.)

13. Sep 23, 2009

### rustyphysics

You are correct Doc Al! The force measured is the spring force and I forgot to take into account the initial velocity... really rusty. How do we take into account the initial velocity?

The set up is essentially like this:

---------- (Rigid Anchor)
| (connectors)
z ("Spring")
| (connectors)
O (Mass - dropped, i.e. with an initial velocity before spring was stretched)

Raw data:
t=Time (ms)
dt = Time step (ms) [each time step is 1ms]
Fi=Force at time step i (kN)
M=mass (kg) [This is a fixed mass for each test]

Other variables:
ai = acceleration (m/s2) at time step i
vi = velocity (m/s) at time step i
si = displacement (m) at time step i
Wi = Work done (J) at time step i

For each time step:
ai = (Fi x 1000 / M) - g
vi = area under a - t graph + initial velocity = 0.5 (ai + ai) * 0.001 + vi+1
si = area under v - t graph = 0.5 (vi + vi-1) * 0.001
Wi = Fi x si

Are the equations in red correct?

I agree this is becoming too engineering, but it has been pretty useful to me so far. Thanks a lot!