Solving Spring System: Box Movement & Total Work

[Huskies213]

Can anyone help me with this i have no idea how to solve it.

You have two identical springs, each with a relaxed length of 50 cm and a spring constant of 450 N/m, they are connected by a short cord of length 10 cm. The upper spring is attached to the ceiling. A box that weighs 98 N hangs from the lower spring. Two additional cords, each 85 cm long, are also tied to the assembly, and they are limp.

A.) How far does the box move?
B.) How much total work do the two spring forces (one directly, the other via a cord) do on the box during that move?

 

[mrjeffy321]

The force a spring will exert when it is streched is equal to the spring constant * the strech distance,
F_s = k*x

The work done on the spring in order to strech it out that distance, x, equals the spring constant * the distance squared,
W_s = k*x^2

If we allow the box to be lowered slowly down to the point of equilibrium (ie. no bouncing up and down, instead a slow and steady lowering until the force upward exerted by the spring = the force downward exerted by gravity), then I think each sping would strech about 11 cm each.

We know the force of the spring(s) up must equal the force of the weight down for the system to be in equilibrium, ad not bouncing around. Since I think we can assume that the ropes and springs are massless, each spring should strech out equal amounts since they both have the same spring constant. Therefore,
F_s = 2*x*450 N/m = weight = 98 Newtons
so, 900x = 98, solving for x, I get a strech distance of .109 meters (10.9 cm) for each spring.

I don't understand the point of the two extra, limp, 85 cm ropes also attached to the system.

The work done by gravity would equal the work done by the springs when the box reaches its lowest point since we will assume that the box is no longer moving. It had some potential gravitational energy, but this was converted into potential spring energy as it decended.
if the box decended ~22 cm (2*11 cm), then it lost 21.56 Joules, meaning that the springs did that much work to counter act this. We can check this answer by plugging back into the spring work equation,
W_s= kx^2
W_s = 450*.22^2 = 21.78 Joules, which is close enough to 21.56 Joules for me, it checks out.

 

[kyle8921]

I can offer you some guidance on how to approach this problem. Firstly, it is important to understand the principles of springs and their behavior. A spring is a type of elastic object that can store mechanical energy when stretched or compressed. The amount of force a spring exerts is directly proportional to the distance it is stretched or compressed, and this relationship is described by Hooke's law, which states that the force is equal to the spring constant multiplied by the displacement from its relaxed length.

To solve this problem, you can use the principles of Hooke's law and conservation of energy. In this case, the box is initially at rest and will begin to move downwards as the lower spring is stretched due to the weight of the box. The upper spring will also be compressed due to the movement of the lower spring, and the cords will remain limp as they are not under tension.

To find the distance the box moves, you can use the equation F = kx, where F is the force exerted by the spring, k is the spring constant, and x is the displacement from the relaxed length. In this case, the total force exerted by the two springs on the box is equal to the weight of the box, which is 98 N. Therefore, you can set up the equation 98 N = (2)(450 N/m)(x), where x is the distance the box moves.

To find the total work done by the two spring forces, you can use the equation W = 1/2kx^2, where W is the work done, k is the spring constant, and x is the displacement. In this case, you will need to calculate the work done by each spring separately and then add them together to get the total work done.

I hope this helps guide you in solving the problem. It is important to understand the principles and equations involved and to carefully set up and solve the problem step by step. If you are still having trouble, it may be helpful to seek assistance from a tutor or classmate who may be able to offer further guidance.