# Work done by spring force and by gravity?

#### Norb_Biderman

I'm reviewing a concept on a system where a mass is hanging from a vertical spring in the presence of gravity. I'm attempting to validate my understanding of conservation of energy when the mass is allowed to slowly extend from its unstretched point to its equilibrium point where the forces cancel out such that there is no kinetic energy at the end point of the movement.

My physics textbook indicates a fundamental equation Wa = -Ws where Wa is the work done by the applied force (gravity in this case) and Ws is work done by the spring force, provided the kinetic energies of the start and end points are zero.

However, I'm confused since this relation seems to be violated. Others on the forum have pointed out that to bring the spring to equilibrium, the work done by gravity is twice as great as that of the spring, as worked out below.

Setting the potential reference level of zero at the unstretched point of the spring (x=0),

Work done by gravity: W = m*g*x

Work done by spring: W = -1/2 * k * x^2

Equating the equations to each other at equilibrium where the net force is zero, we get:

2 * m * g = -k * x

How can this be? Does this violate the work expression I wrote several paragraphs above? Physically, shouldn't the work done by gravity equal to the work done on the spring? I've pored over explanations provided by others on this forum, but I've found them lacking for some reason.

Any insight would be greatly appreciated!

Last edited:
Related Classical Physics News on Phys.org

#### Fewmet

You wrote that the work done by the spring is
W = -1/2 * k * x
But kx is the force applied by the spring. Work is
W = -1/2 * k * x2

#### Norb_Biderman

My bad. That's what I meant. Edited.

My confusion still stands.

#### rcgldr

Homework Helper
When the net force is zero, the mass would still be moving downwards due to it's momentum. You'd need to take into account the kinetic energy of the mass at the moment the forces cancel.

#### Doc Al

Mentor
Physically, shouldn't the work done by gravity equal to the work done on the spring?
No. Note that the force of gravity is greater than the restoring force of the spring as the mass falls from the unstretched position to the equilibrium position. While the restoring force varies from 0 to mg (at equilibrium), the gravitational force remains constant at mg. The additional work done by gravity gives the mass kinetic energy (as stated by rcgldr).

Compare the resulting final energy of the spring-mass system in this situation, where you let the mass fall, to a situation where you gently lower the mass.

#### Norb_Biderman

No. Note that the force of gravity is greater than the restoring force of the spring as the mass falls from the unstretched position to the equilibrium position. While the restoring force varies from 0 to mg (at equilibrium), the gravitational force remains constant at mg. The additional work done by gravity gives the mass kinetic energy (as stated by rcgldr).

Compare the resulting final energy of the spring-mass system in this situation, where you let the mass fall, to a situation where you gently lower the mass.
In the scenario where the weight is abruptly let go of when it's attached to the spring, I understand what is happening, but in the case where we gently lower the mass, I cannot reconcile the observation mathematically. Our hand is working against the motion of the weight, therefore we are doing negative work on it. How do I quantify that work? Would it be considered a position-dependent force?

My original question centers around the original equation Wa = -Ws. In full, it is written out as:

$\Delta$K = Kf - Ki = WA + WS where WS = 1/2 k*x2

Gently lowering the mass to its equilibrium point is equivalent to initial and final kinetic energies being zero. In this case, I am unable to reconcile the idea of energy conservation.

If I am still being unclear, refer to this thread:

Last edited:

#### Doc Al

Mentor
In the scenario where the weight is abruptly let go of when it's attached to the spring, I understand what is happening,
Just to be clear: Do you recognize that this is the situation you described in post #1? Does that mean you now understand the issue there?
but in the case where we gently lower the mass, I cannot reconcile the observation mathematically. Our hand is working against the motion of the weight, therefore we are doing negative work on it. How do I quantify that work? Would it be considered a position-dependent force?
Think of it this way: When you gently lower the weight, your hand exerts just enough force so that the net force on the mass is zero (to a good approximation). The force of gravity (mg down) + force of your hand (up) + spring restoring force (kx up) = 0.

My original question centers around the original equation Wa = -Ws. In full, it is written out as:

$\Delta$K = Kf - Ki = WA + S where WS = 1/2 k*x2

Gently lowering the mass to its equilibrium point is equivalent to initial and final kinetic energies being zero. In this case, I am unable to reconcile the idea of energy conservation.
When the mass is gently lowered, the applied force (gravity + hand) is exactly equal and opposite to the spring force, so the work done by the applied force will equal the spring potential energy.

In your original question, there was no hand gently lowering the mass, so the work done by gravity was greater than the spring PE, giving the mass KE.

If I am still being unclear, refer to this thread:

I don't quite understand the relevance of that thread.

#### Norb_Biderman

I apologize for the confusion. In my original post, I meant gently lowering the mass to its equilibrium point using our hand. I used the term "slowly." That idea probably could be interpreted as physically ambiguous. Should have been more clear about that.
Just to be clear: Do you recognize that this is the situation you described in post #1? Does that mean you now understand the issue there?
Yes, we are on the same page now.
Think of it this way: When you gently lower the weight, your hand exerts just enough force so that the net force on the mass is zero (to a good approximation). The force of gravity (mg down) + force of your hand (up) + spring restoring force (kx up) = 0.

When the mass is gently lowered, the applied force (gravity + hand) is exactly equal and opposite to the spring force, so the work done by the applied force will equal the spring potential energy.
Makes sense.

I don't quite understand the relevance of that thread.
In that thread, the original poster used two different expressions, to quantify the energy change, the expression for gravitational potential energy and the energy relation derived from Hooke's Law, and got two different answers.

However, I now see that the gravitational potential energy expression used by the original poster in that thread did not account for the work done by the hand gently lowering the weight to its equilibrium position. Is this correct?

#### Doc Al

Mentor
In that thread, the original poster used two different expressions, to quantify the energy change, the expression for gravitational potential energy and the energy relation derived from Hooke's Law, and got two different answers.
Ah, I see now.
However, I now see that the gravitational potential energy expression used by the original poster in that thread did not account for the work done by the hand gently lowering the weight to its equilibrium position. Is this correct?
Exactly. The work done by gravity - work done by hand = increase in spring PE.

• Limmin

Appreciate it!

### Physics Forums Values

We Value Quality
• Topics based on mainstream science
• Proper English grammar and spelling
We Value Civility
• Positive and compassionate attitudes
• Patience while debating
We Value Productivity
• Disciplined to remain on-topic
• Recognition of own weaknesses
• Solo and co-op problem solving