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  1. Mar 21, 2006 #1
    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?
  2. jcsd
  3. Mar 21, 2006 #2


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    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 dont 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.
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