# Work done by spring problem

The problem is hard to describe, but ill give it a shot:

there is a 50cm long spring ( k = 500 N/m) hanging from the ceiling and a 10cm cord connecting it to another identical spring, which supports a 100N box. In addition, there are two (limp) 85cm cords. One is hanging vertically from the celing and is connected to the lower spring. The other one is hanging from the top spring and is connected to the 100 N box. If the 10cm cord in the middle is cut, I need to find out if the box moves up or down, how far the box moves, and the total work the spring forces do during the move.

Supposedly, the answers to the first two are up and 5cm. However, I am clueless as to how to begin this problem. Even when I draw all the forces in, nothing seems to make sense. I know it's a difficult problem, but I dont even know how to tell that the box is moving up.

OlderDan
Homework Helper
Maybe I am not seeing the geometry correctly, but I can't imagine the initial move being upward. I would try an energy approach. It sounds to me like both springs will be supporting the same weight the moment the secondary strings become taught and will stretch by the same amount.

yeah, I have in a free body diagram (finally) a tension force exerted on one spring [upwards], and a spring force exerted on the block [upwards] (for the spring attached to the block) and a spring force exerted upwards with a tension force exerted downards, but because of newton's 3rd law the tension on the rope is also pulling upwards on the block.

However, I dont see how to approach this as an energy problem since no speeds are given and I dont know how far the force acts on the block (after all I have to find the displacement in the second part).

OlderDan
Homework Helper
yeah, I have in a free body diagram (finally) a tension force exerted on one spring [upwards], and a spring force exerted on the block [upwards] (for the spring attached to the block) and a spring force exerted upwards with a tension force exerted downards, but because of newton's 3rd law the tension on the rope is also pulling upwards on the block.

However, I dont see how to approach this as an energy problem since no speeds are given and I dont know how far the force acts on the block (after all I have to find the displacement in the second part).
The system starts from rest. If the block falls when the string is cut, it loses gravitational potential energy (GPE) and gains kinetic energy (KE) and accelerates until the secondary strings are taught and the springs start to stretch. A that point, the kinetic energy diminishes and spring energy (SE) increases.

The springs may have more energy before the string is cut than they do at any later time. I still don't see an initial upward move, but the springs will be shorter after the string is cut so I'm not saying it can't happen for sure. What is certain is that the total energy when the mass comes to rest is potential energy, and the total energy is conserved.

I solved the problem with Hooke's Law, with series and parallels. The second question is actually necessary to answer the first.

OlderDan
Homework Helper
I solved the problem with Hooke's Law, with series and parallels. The second question is actually necessary to answer the first.
The energy calculation says it moves down 39.5cm before stopping, assuming I interpreted the problem correctly. I assume the 50cm springs are already strecthed initially by W/k = 20cm, so the equilibrium length is 30cm.

so...since both springs are stretched by the block, the restoring forces on the springs when the rope is cut causes the block to move upward? (i.e., back to equilibrium?).

So then do the combined potential energies of the two springs equal the change in potential energy (5 J), thus giving the distance raised of 5cm?

I don't know if I am correct in my assumptions, which is why I am asking these types of questions in this post.

OlderDan
Homework Helper
so...since both springs are stretched by the block, the restoring forces on the springs when the rope is cut causes the block to move upward? (i.e., back to equilibrium?).

So then do the combined potential energies of the two springs equal the change in potential energy (5 J), thus giving the distance raised of 5cm?

I don't know if I am correct in my assumptions, which is why I am asking these types of questions in this post.
Here is how I see the problem from your description. I don't ever see this 5cm rise being a particularly interesting point, even if the box does get there. The initial configuration is the 100N box at rest at the bottom of a spring (k = 500N/m) of length 50 cm hanging from a string of length 10cm connected to an identical spring above that is connected to the ceiling. The box is initially 110cm below the ceiling. Each spring is elongated 20cm by the 100N force [100N/500N/m = 0.2m]. The relaxed length of each spring is therefore 30cm. Without the box, the bottom of the second spring would be 70 cm below the ceiling. The series combination of the two springs has an effective spring constant of 100N/.40m = 250N/m.

When the string is cut, the relaxed positiion of the bottom of the top spring is 30cm (as before) with 85cm of string hanging down to the box and the bottom of the second spring is at the level of the box with 85cm of string above the spring. The relaxed level of both supports is 85cm + 30cm = 115cm from the ceiling. The weight of the 100N box is shared equally by the two springs, each supporting 50N of weight and stretching 10cm at equilibrium. The effective spring constant of the two springs in parallel is 100N/0.1m = 1000N/m. If the box were hung at rest it would be 125cm from the ceiling. Of course it is not at rest when it reaches this level.

You can choose the zero of GPE anywhere you want. If you choose the ceiling, the GPE will always be negative, but it is a convenient choice, so I used it. The total energy of the system is the inital GPE plus the energy of each spring elongated by x = 20cm (or the effective spring elongated by 2x = 40cm; either way gives the same spring energy)

E = ½kx² + ½kx² + Mgh_o = (500N/m)(0.2m)² - 100N(1.10m) = 20J - 110J = -90J

When the connecting string is cut, the box will find new positions with this much energy, most of which involve some amount of kinetic energy. But at the lowest point
the energy will be all potential. The springs will both be stretched by an amount y from the relaxed length of 30cm. At this point the total energy will be

E = ½ky² + ½ky² + Mgh_f = (500N/m)(y²) - 100N(1.15m + y) = -90J

Rearranging terms gives the quadratic equation

0 = (500/m)(y²) - 100y - 25m

The solution is

y = {[1 + sqrt(6)]/10}m = 34.5 cm (the negative root solution is unphysical. Why?)

The box stops 34.5cm below the new relaxed postion of the springs, or 115cm + 34.5cm = 149.5cm below the ceiling. The energy of the springs at this point is 59.5J which is 39.5J more than they started with. Where did this additional energy come from? It came from the loss of GPE which is now -149.5J compared to the initial -110J.

Now the springs are strethed well below the system equilibrium point. The box reverses position and heads upward. How high do you think it goes and why?

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I just realized I made a mistake...the 50cm is the relaxed length of the spring. I apologize for not reading my question over carefully enough.

OlderDan
Homework Helper
I just realized I made a mistake...the 50cm is the relaxed length of the spring. I apologize for not reading my question over carefully enough.
OK That will change things. My approach is still valid, but the answer could be quite different.

I can see a 5cm distance in the problem, but it is not relative to the starting point of the box. It is relative to the new height where the springs are relaxed.

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OlderDan