Department of Energy Argonne National Laboratory Office of Science NEWTON's Homepage NEWTON's Homepage
NEWTON, Ask A Scientist!
NEWTON Home Page NEWTON Teachers Visit Our Archives Ask A Question How To Ask A Question Question of the Week Our Expert Scientists Volunteer at NEWTON! Frequently Asked Questions Referencing NEWTON About NEWTON About Ask A Scientist Education At Argonne Spring System Energy
Name: Shawn D.
Status: student
Age: 16
Location: N/A
Country: N/A
Date: 2/26/2003

My physics teacher found a discrepancy in our book. Say you are given a vertical spring with a mass hanging from it. If the weight is displaced downward, and let go, the book says that the potential energy of gravity can be ignored when finding the weight's maximum velocity. It factors in the change in the spring's potential energy, but disregards gravity. Please help us restore our faith in the book we invested so much work on.

The book is correct, but misleading. The condition of MAXIMUM velocity is the key phrase and the resolution of the apparent paradox. Without going through the derivation, the maximum velocity occurs when the spring passes through its equilibrium position. For that instant of maximum velocity the spring is in its equilibrium so the force of the spring Fs = -k(X - Xo) = 0 just balances the gravitational force Fg = mgh. All of the energy is in the form of kinetic energy. I do agree that the book should not have just left such a remark unexplained.

Vince Calder

Your book is absolutely right, even though it seems to violate common sense when you first see it. I think I might be able to convince you with this argument:

We all know that Newton's Laws rules the roost here, so F = ma. Consider the F. If you consider y to be measured upward from the position of the mass when the spring is not extended (the mass is not yet attached), then

F = -ky - mg.

Here k is the spring constant and mg is the weight of the mass. Now do a change of variable to, say, z = y + d, where z is measured from the mass position when it is hanging on the spring and is at rest. then

F = 0 = -kd - mg and d = -mg/k so

y = z - d.

Notice that then F = -ky - mg = -kz + kmg/k - mg = -kz. So if you measure the displacement of the mass from its equilibrium position, the effect of gravity vanishes! What really happens, of course, is that the extra spring tension just exactly balances the weight of the mass.

Best, Dick Plano, Professor of Physics emeritus, Rutgers University


The effect of gravity is take into account by your choice of the zero point for the spring. If you choose the zero point to be the point of equilibrium, the spring force you are using contains the effect of gravitational force. The spring potential energy contains the effect of gravitational potential energy. You have essentially fused them into one force equation. If instead of equilibrium, you choose the zero point to be where the spring is when it has no weight hanging from it, then you must include gravitational potential energy.

Dr. Ken Mellendorf
Physics Instructor
Illinois Central College

I think the idea that the book intends to get across is that you can choose arbitrarily the height to which you will assign a gravitational potential energy of zero. You can do this because you're only interested in changes in gravitational potential energy.

However, to calculate the maximum kinetic energy that will result from a given spring extension, you cannot ignore the change in gravitational potential energy associated with that spring extension. If the spring force at the given extension is large compared to the weight of the mass, then you can neglect gravitational potential energy and get approximately the right answer, but the exact answer has to include it.

Tim Mooney

While it may not seem intuitive at first, the book is correct, the maximum velocity is not affected by gravity. This isn't to say that gravity has no affect on the system, only that its effect on the system does not affect the velocity.

Recall that the force due to gravity is m * g, or mass times the acceleration due to gravity (9.8 m/(s*s)).

Recall also that force from the spring is -k * x, or the spring constant times the displacement.

In its initial state, this spring system has no acceleration, so the total forces must be zero:

-m*g - k * x = 0

So we can solve this for x:

m * g / k = -x

We will call this initial displacement Xg, or the displacement due to gravity, and its sign will be negative since it is a downward displacement.

Now, when you pull the spring downward, you will increase the force acting upward on the spring. We will call this displacement Xd, and its sign will be negative, since it is a downward displacement.

So when you have displaced the system, the total force will be:

-m * g + k * Xg + k * Xd = Flowest

And, at the highest point, the forces will be:

-m * g + k * Xg - k * Xh = Fhighest

Since the energy at the top must be the same as the energy at the bottom, and since at the time the they are at the top and bottom, the mass is not moving, all the energy is potential at these points.


-m * g + k * Xg + k * Xd = -m * g + k * Xg - k * Xh

Now we simplify. First we already know that -m*g + k * Xg = 0, so:

k * Xd = -k * Xh

And we can stop there and realize that the energy contributed by gravity is constant throughout, and its only effect is to displace the entire system downwards by a fixed amount. In other words, the displacement of the spring by gravity reduces the potential energy due to gravity of the entire system throughout its entire existence. So the mass is lower at its lowest point that it otherwise would be, and it is lower at its highest point than it otherwise would be.

I hope this helps,
--Eric Tolman


Far be it from me to argue that text books never contain errors! Despite the author's best efforts they do get there so it is always a good practice to seek to understand the subject (as you and your classmates are) rather than to accept information blindly.

In this case it sounds like a problem that you have been asked to work or that is being presented as an example. Sometimes authors like to simplify problems by removing factors from consideration. This is how a lot of scientific work is done. We start by assuming certain factors do not have a big impact on our results and come up with a theory that explains the major effects, then we start finding the factors that result in minor corrections until the theory has enough components to fully explain the observations.

Now, in this case the change in potential energy due to gravity has been neglected. In essence the author is assuming (or asking you to assume) that the spring force constant is much greater than the gravitational force. Have you worked the problem both ways? How much difference does the gravitational force make? Was the author justified in neglecting gravitational potential energy?

Greg Bradburn

Click here to return to the Physics Archives

NEWTON is an electronic community for Science, Math, and Computer Science K-12 Educators, sponsored and operated by Argonne National Laboratory's Educational Programs, Andrew Skipor, Ph.D., Head of Educational Programs.

For assistance with NEWTON contact a System Operator (, or at Argonne's Educational Programs

Educational Programs
Building 360
9700 S. Cass Ave.
Argonne, Illinois
60439-4845, USA
Update: June 2012
Weclome To Newton

Argonne National Laboratory