- #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
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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