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

In summary, the work done in lifting an object with a spring will be greater than lifting it with a rod due to the additional work required to stretch the spring. However, the apparent weight of the object will remain the same, regardless of whether it is lifted with a spring or a rod. The amount of stretching of the spring depends on the Hooke's constant, with a weaker spring requiring more work to be done. The acceleration and timing of the lift also play a factor in the perceived weight of the object.
  • #1
Mr Virtual
218
4
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
 
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  • #2
Yes, you are correct.
 
  • #3
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
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
Mr Virtual said:
Also, will we feel the object to be heavier, lighter or equal to its actual weight when we lift it with a spring?


Only if the rod is of a different (heavier, lighter)/same (equal) weight than the spring.
 
  • #6
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
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.)
 

1. How does the work done in lifting an object with a spring compare to lifting it with a rod?

The work done in lifting an object with a spring depends on the spring constant and the distance the spring is stretched. If the spring is stretched a greater distance, more work is done compared to lifting the object with a rod. However, if the spring constant is very low, the work done may be less than that of lifting with a rod.

2. Does the weight of the object affect the work done when lifting with a spring?

Yes, the weight of the object does affect the work done when lifting with a spring. The greater the weight of the object, the more work is required to stretch the spring and lift the object. This is because the force required to stretch the spring increases with the weight of the object.

3. How do friction and air resistance impact the work done when lifting with a spring?

Friction and air resistance can have a small impact on the work done when lifting an object with a spring. Friction may slightly decrease the work done as it provides a resistive force against the motion of the object. Air resistance may also have a small effect, but it is usually negligible unless the object is very large or moving at high speeds.

4. Can the work done in lifting an object with a spring be calculated?

Yes, the work done in lifting an object with a spring can be calculated using the formula W = 1/2kx^2, where W is the work done, k is the spring constant, and x is the distance the spring is stretched. This formula assumes that the spring is being stretched from its equilibrium position and no other external factors are involved.

5. Is there a limit to how much work can be done when lifting an object with a spring?

Yes, there is a limit to how much work can be done when lifting an object with a spring. This limit is determined by the maximum amount of force the spring can exert and the distance it can be stretched. If the object is too heavy or the spring is stretched beyond its limit, the work done will decrease and the spring may even fail.

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