Why is a spring a conservative force?

[OmegaKV]

Let's say you are holding a bowling ball in the air, and there is a spring on the ground directly below it. You drop the bowling ball and it lands on the spring, compressing it, and then the bowling ball rolls off the spring onto a point P right next to the spring. Wouldn't this do a different amount of work than if you were to just throw the bowling ball directly onto P, bypassing the spring?

 

[Orodruin]

You will have to be more specific. What would do a different work on what?

 

[drvrm]

Wouldn't this do a different amount of work than if you were to just throw the bowling ball directly onto P, bypassing the spring?

i agree with the above comment -that more details are reqired
but as it stands the amount of work will be different as the paths are different and we do not know what type of forces are operating in the next throw- as far as spring action is concerned if it is ideal spring the motion will be holding the conservation of energy.

 

[OmegaKV]

You will have to be more specific. What would do a different work on what?

The work done on the bowling ball by gravity and the spring.

Here are two ways to get to P with ball initially at height h:

Method 1 (bypass the spring):
Step 1) Move the ball horizontally so that it is positioned above P (0 work)
Step 2) Drop the ball (Work = mgh)
Total work on ball = mgh

Method 2 (drop ball on spring):
Step 1) Drop the ball, and the ball lands on the spring. If the uncompressed spring is y tall, the work from this step is: mg(h-y) plus the integral of (mg-ky) from 0 to y.
Step 2) Move the ball horizontally onto P (0 work if floor is frictionless).
Total work on the ball = mg(h-y)+mgy-(1/2)ky^2=mgh-(1/2)ky^2

 

[Orodruin]

You are missing the work done by the floor on the ball in order to stop it in the case where it is not dropped on the spring. There is nothing strange going on here.

 

[OmegaKV]

You are missing the work done by the floor on the ball in order to stop it in the case where it is not dropped on the spring. There is nothing strange going on here.

What if the spring is tight enough that the ball never completely hits the ground, and then you push the ball off the spring right when the ball's velocity reaches 0 (i.e. right before the ball starts going back up)?

 

[Orodruin]

What if the spring is tight enough that the ball never completely hits the ground, and then you push the ball off the spring right when the ball's velocity reaches 0 (i.e. right before the ball starts going back up)?

This does not change the fact that the force from the floor will do work in the other case.

 

[mfig]

Wouldn't this do a different amount of work than if you were to just throw the bowling ball directly onto P, bypassing the spring?

Either way you are dealing with a generalized spring. Take the case where the ball is dropped directly on to the floor. If the floor is undamaged and left without plastic deformation, then it too acted as a spring that stopped the ball. Real materials, like a floor, have some level of elasticity. A floor simply has a different spring constant (elastic modulus) than the actual spring.

 

[OldYat47]

In Method 1 the rebound of the spring must be accounted for. In Method 2 the floor never does any work unless it deflects (work = force X distance). For that calculation you would need the floor stiffness characteristics (another spring!). In the end, the available energy is in mgh. Without getting into the weeds and tiny details the end result has to add up to mgh for conservation of energy.