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Homework Help: Maximum distance

  1. Oct 21, 2007 #1
    A 0.150-kg frame, when suspended from a coil spring, stretches the spring 0.050 m. A 0.200-kg lump of putty is dropped from rest onto the frame from a height of 30.0 cm.

    Find the maximum distance the frame moves downward from its initial position in meters.

    First I found the spring constant, k, which is given by F = -kx.

    F = -kx
    k = -F/x = -(0.150kg *9.8 m/s2)/(0.050m) = 29.4 N/m

    Next, I found the force that the lump of putty makes on the frame. The force is given by F = ma. The mass of the putty, 0.200kg, and the acceleration which is due to gravity, -9.8m/s2.

    F = ma = 0.200kg * -9.8m/s2 = 1.96 N

    Now the spring constant, k (29.4N/m), and the force of the putty, 1.96N, so I finally solve for the distance the frame moves.

    F = -kx
    x = -k/F = -(29.4N/m)/(1.96N) = -15m

    It is negative, because it moves in the -x direction.

    So, the frame moves 15m downwards is what I get but my program says it is wrong. Can someone please help me?
  2. jcsd
  3. Oct 21, 2007 #2


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    Homework Helper

    This part looks OK:

    For the next part, the 1.96 N is not the force that the putty exerts on the frame, since the putty was released 0.3 m above the frame. You can consider conservation of energy to get the distance the spring will go to bring the putty to a stop.
  4. Oct 21, 2007 #3
    Ok. so how about:

    1/2mv^2= 1/2 kx^2

    F= - k x
    where F= mg = (0.150-) (9.80m/s2)
    F = 1.47 N
    1.47 N= -k ( - 0.050 )
    k = 1.47N/0.050m
    k = 29.4N/m
    v22 = v12 - 2gh, where v1 = 0 and h =- 30 cm = - 0.30m
    v22 = - 2(9.80m/s2)(- 0.30m)
    v22 = 5.88
    v2 = 2.425 m/s
    Now, plugging these values in I get
    mv2 = k x2
    x2 = mv2/k = (0.200)(5.88 m2/s2)/(29.4N/m)
    x2 = 0.04
    x = 0.2 m

    So the maximum distance is:
    X = 0.05 m + 0.20 m
    X = 0.25 m

    Is this right?
  5. Oct 21, 2007 #4


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    Homework Helper

    Looks OK to me.
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