# Is work done in lifting an object with a spring more than lifting it with a rod?

1. Jun 14, 2007

### Mr Virtual

Hi everyone

Please note: It is not a homework-type question. I haven't got springs in my course.

Suppose we have a rod, a spring and a heavy object of weight 50 N. If we lift the object with the help of a rod upto 1m, work done by us is 50*1=50 J.

Now we remove the rod and lift the same object with a spring. Will the work done in lifting it to the same height be greater or less than the work done in lifting it with a rod?
Also, will we feel the object to be heavier, lighter or equal to its actual weight when we lift it with a spring?

For a spring,
F = kx
W = 1/2 * k * x^2

(I know all of you know these simple formulae. Just for those who might have forgotten it for a short while)

As far as I know, when a spring is used in lifting, it first stretches until its restoring force becomes equal to the weight of the object (50 N). The amount of stretching depends on k (Hooke's constant).
After that the spring exerts same force (50 N) on our hand, just like in the case of lifting with a rod. The difference is just that this time work done is greater because, besides lifting the object to 1m, we have also had to do work in stretching the spring. Lesser the value of k, more will be the work done in stretching.
According to me, apparent weight of the object will remain same (50 N), no matter whether we lift it with a spring or a rod.
Am I right?

I know this is a stupid question, but I just wanted to assure myself.

Thanks
Mr V

Last edited: Jun 14, 2007
2. Jun 14, 2007

### Staff: Mentor

Yes, you are correct.

3. Jun 15, 2007

### Jack111

The work done AGAINST GRAVITY will be the same. You will need to do work on the spring to extend it to the point that it exerts a 50N force on the weight, after which point you will lift the weight normally. The work that you've done on the spring is recoverable later since it's stored as potential energy in the spring.

To lift the weight, F = kx = Mg
=> x = Mg / k

W(spring) = 1/2 * k * x^2 = 1/2 * k * (Mg/k)^2
= 1/2 * (Mg)^2 / k

A rod is of course the limit of a spring as k-> infinity, and we can see here how in this case the work that you have to do against the spring tends to zero in this limit.

4. Jun 15, 2007

### Staff: Mentor

If you attach an object to a compressed spring and release it, you end up with simple harmonic motion and the extra force of the spring is absorbed in the acceleration of the object.

Lifting a 50n object with 50n of force imples an arbitrarily small acceleration and long time to lift the object.

5. Jun 15, 2007

### rewebster

Only if the rod is of a different (heavier, lighter)/same (equal) weight than the spring.

6. Jun 15, 2007

### Staff: Mentor

Actually, if we're supposed to consider the case in my post, the object will feel heavier when lifted with a spring if it is lifted faster than with the rod.

7. Jun 15, 2007

### rewebster

Then it also depends on the spring and the timing. A very weak spring will extent until it is a straight 'rod' (depending on the weight of the 'weight'), a very stiff spring will act act like a rod (depending on the weight of the 'weight'); if the weight is lifted fast on a 'medium' spring and you measure the 'feel' of the 'weight' of the object at the point where it 'could' feel weightless or even a 'negative' weight. (There's too many unknown variables.)