# What is the work done.

A weight of mass m is connected to a spring of spring constant k. Initially the weight hangs down and is at rest. Support the weight by hand and lift it up slowly until the length of the spring is equal to its natural length. What is the work done by the hand during this process? Let the gravitational constant be g.

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Hootenanny
Staff Emeritus
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Please show some working or intial thoughts...

~H

Well what I am trying to get here is the work done. So I will try to multiply the constant k to the mass m and the gravitational g. But m*g*k is not on the choices. Maybe this is not an easy problem though.

HINT: Try to obtain an expression for the extension of the spring when the weight was initially at rest.

What I have in mind is m*g*k to get the work done. But when the object is pulled upwards to get its natural length, then that is another story. So the formula will be different and thats what I am trying to figure it out.

k is a spring constant and its unit is N/m, so the product mgk will not get you an expression for work done. (For one, the unit of the product mgk does not correspond to the unit of work, the Joule)

The hint in my previous post is probably a neccesary step in your working, so why not start there?

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Doc Al
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willydavidjr said:
What I have in mind is m*g*k to get the work done.
Not sure what you're thinking here. "k" is the Hooke's law spring constant, so muliplying that times the weight doesn't make sense. (What are the units of work? What is Hooke's law?)

Two ways to go about finding the work done. The easy way is to use energy methods, if you've covered that. If not, then you've got to figure out what force must be exerted on the mass at each point and then integrate over the distance to find the work done. (Hint: Since the mass is lifted slowly, it's always in equilibrium.)

Now what do you think if the work done is $$\frac{2 m^2 g^2}{k}$$?

Hmmm...that's not the answer I got.

Perhaps you would like to show some working?

I have change and verified my solution. What I got now is $$\frac{m^2g^2}{2k}$$. Divide the square of mass and gravity to the potential energy stored on the spring. It is 1/2 in relation to k. The unit is more likely to be N.m. And it is the unit for the work done.

You're way off track here.

The way to approach the problem is to think in terms of the energy.
When the spring is stretched, there is some potential energy there.
When you raise the mass, such that the spring is at its natural length, you are increasing the potential energy of the system.

What is the relationship between these energies?

Clarification: the potential energy in each situation is not the same kind of energy. When the spring is stretched, it's from the spring. When the mass is at its final height, the potential energy is gravitational in nature.

Doc Al
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willydavidjr said:
I have change and verified my solution. What I got now is $$\frac{m^2g^2}{2k}$$.

Divide the square of mass and gravity to the potential energy stored on the spring. It is 1/2 in relation to k.
I assume you are explaining your reasoning behind your answer, in which case I don't know what you mean. The way to get the answer is to consider the change in energy as the mass is lifted. (First you have to figure out how far the mass must be lifted.) Both spring and gravitational potential energy will change as the mass is lifted; the net change in energy will equal the work done by the hand.

In standard units, the answer will be in N-m, otherwise known as Joules.

Sorry. But on my question, there are options like what I have shown. So I consider one of the options just like my answer based on the standard units I can get. My answer above will generate in Joules.

Doc Al
Mentor
So... you are picking the answer out of a set of choices based on which answer gives the correct units? That's OK as a test taking strategy (if you are lucky enough that only one of the answers has the correct units for energy), but this problem is simple enough that you should be able to solve it directly.

"Elastics"

Well since a weight is hung to a spring and a spring has elasticity the weight will cause the spring to extend beyond its natural length, call that extension 'x'.

Now think of the forces on the weight, the tension of the spring due to its elasticity acting upwards, call that 'F' and the gravitational pull or weight, call that 'W'.

Now u should probably know that Elastic force = Spring constant x Extension
That is, F=kx
Now this force is balanced by the weight, W and u probably know that W=mg
Therefore, F=W
kx=mg
x=mg/k

Before the weight was taken up by the hand, it had elastic potential energy. Now when the weight was taken up that elastic potential energy should be converted to gravitational potential energy since it has gained height with respect to that level (supposing we consider the lowest point of the weight when hanging freely from the spring as ground level for g.p.e)

So the work done by the hand must be equal to the g.p.e since it no longer has e.p.e as the spring is back to its natural length.

g.p.e=mgh
=mg x mg/k (here h is the extension of the spring)
(Now here i m really facin a problem to get the superscripts properly so da rest is for ya to find out and now dont tell me u cant do that........):tongue2:
***B***

Doc Al
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reincarnated_soul said:
So the work done by the hand must be equal to the g.p.e since it no longer has e.p.e as the spring is back to its natural length.
Not exactly. The work done by the hand equals the change in total potential energy (gravity & elastic). (The elasticity makes it easier to lift, so less work is needed.)

Thanks a lot reincarnarnated_soul. To get thats superscipts, get a reading about latex. It is worth reading and you'll probably get a long run on this forum.

But I have a different approach to this problem.

If we get x=mg/k, x here will be the distance or h in your answer.
Then to get the work done: W=Fd
Force=mg
X=h=mg/k

Then substituting the variables
W=mg * mg/k
$$W=\frac{m^2g^2}{k}$$

Doc Al
Mentor
willydavidjr said:
But I have a different approach to this problem.

If we get x=mg/k, x here will be the distance or h in your answer.
Then to get the work done: W=Fd
Force=mg
X=h=mg/k

Then substituting the variables
W=mg * mg/k
$$W=\frac{m^2g^2}{k}$$
This answer is incorrect. Note that the force that the hand exerts is not constant; it varies from 0 to mg. (Hint: Find the average force exerted by the hand.)

Well, I'm curious since this problem wasn't finally solved. I'm trying to solve it myself but I have a problem.

Ok, so my approach, as suggested, was doing it with energy and work. Work is the difference between the initial and final potetial energy of te system, right?.

So, at the beggining, when the spring was streched, we only had elastic potential energy, which was Ui = (kx^2)/2 ---> x equals the streched distance of the spring.
At the end, we had only gravitational potential energy, since the spring is in its initial position. This time this energy is Uf= mgx (the 'height' of the mass is x).

since its easy to know that x= (mg)/k, i can make the difference between the initial and final energies...

Ui-Uf= (k(x)^2)/2 - mgx = 1/2*k(mg/k)^2 - mg (mg/k) = (mg)^2/2k - (mg)^2/k = -(mg)^2/2k = W

Is this reasoning correct? I'm kinda confused about the sign (negative) of the work...

work is the difference between the final and initial potential energy of the system (although this only applies if the force is conservative), that's why you were getting a negative sign.

Well, thanks for that kaworito...But I looked back from different options for the answer and I found your answer but without the negative sign. I think it is positive.