Work A mass m hangs from a spring.

  • #1
A mass m hangs from a spring. You push up on the mass until the spring reaches its natural length. How far will the hanging mass stretch the spring? How much work do you do against gravity? Against the spring?

When the mass comes to rest, -mg+kx=0 or x=mg/k

The work done by you against the spring should be zero, because you act in the same direction as the spring? I don't get this part in particular. What does it mean, "work against a force?"

The work done against gravity is just the force required to balance gravity integrated wrt x. kx-mg+F=0 or F=mg-kx. W=[inte](mg-kx)dx=mgx-.5kx2|x0=mgx-.5kx2

Alternatively, I wrote W=dK=0=.5kx2 + Wyou -mgx
or Wyou=mgx-.5kx2=mg(mg/k) - .5k(mg/k)2

Is this ok? My main confusion is the idea of work against a force.
 
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Answers and Replies

  • #2
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It does seem like a poorly worded question. I would assume the mass started at its equilibrium point & I would interpret the work aspects of it the same way you did.

I'm also not sure how to interpret "How far will the hanging mass stretch the spring?"

Is that just asking how far did the mass stretch the spring when it was at equilibrium?

Or does that imply that the mass is to be released after being pushed up to the point of the natural length of the spring? If so, I think it will then fall twice as far before the spring tension stops it & it bounces back up, using
(1/2)ky2 = mgy
y = 2mg/k
(But who knows if this is what they were asking.)
 
  • #3
You are to assume that the mass hangs at rest from the spring so that it stretches the spring a certain length. That was my interpretation of the problem.
My biggest concern was the second part- the "work against" part.
 
  • #4
HallsofIvy
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In order to move a mass "against a force" you must apply an opposing force of a least the same strength. The work done "against the force" is the integral of that force applied over the distance.

You have correctly determined that the vertical spring will hang down a distance x= mg/k.

In order to move it upward, you must oppose gravity- and the spring will help. The gravitational force, downward, is, of course, -mg. The spring is pulling upward with force kx so the net force on the mass is kx-mg (which is 0 when x= mg/k). The force you must apply to move the mass upward (without accelerating it) is the opposite:
mg- kx, exactly what you have. Integrate that from 0 to mg/k (or, more correctly from -mg/k to 0) which is exactly what you did.

You do, I assume, recognize that the answers you got by doing the problem in the two different ways are exactly the same?
 
  • #5
Originally posted by HallsofIvy

You do, I assume, recognize that the answers you got by doing the problem in the two different ways are exactly the same?
Haha... Yes, I do. I just wanted to make sure my first answer was right. My biggest concern was this "work against" idea.

Originally posted by HallsofIvy
In order to move a mass "against a force" you must apply an opposing force of a least the same strength.
IOW, if the sign of the works are different then they are against each other? In this example, gravity is a negative force acting along a positive direction (W<0) and my force is positive acting along a positive direction (W>0). But the spring also has a positive work so I do not work against it. So I do (mg)^2/(2k) J of work against gravity and 0J of work against the spring.
 
  • #6
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Work is defined as the dot product of the force vector by the displacement vector. So your work is the scalar product of YOUR force and the displacement vector, regardless of any other forces that may be involved. Gravity's work is the scalar product of the gravitational force by the displacement vector. In this case, of course, each is the negative of the other.

The work itself is a scalar, not a vector, so it has no direction.

So, Stephen, I share your annoyance at the idea of work "against a force". It seems to me that "against" is nothing but a reference to the relative directions of the forces involved, and they might as well say you are working "with the spring". It would be even more muddled if the forces were not diametrically opposed to each other, or if there were many forces all acting in different directions.
 

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