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Homework Help: Hooke's law problem

  1. May 7, 2008 #1
    A 0.2 kg block of wood is attached to a spring with a spring constant k = 25 N/m. The block is initially at rest and the spring is at its equilibrium length aligned along the x-axis. A dart of mass 0.05 kg is thrown at a block of wood, undergoes an inelastic collision and sticks into the block. the initial speed of the dart is 0.1 m/s and is parallel to the x-axis. The maximum displacement of the block from its equilibrium position in m is: ?

    2. Relevant equations

    3. The attempt at a solution

    I used m1Vi + m2Vi = (m1+ m2) Vf
    to find Vf = 0.1 m/s

    I know F= k x
    I dunno know how you would find F.
    Please help
  2. jcsd
  3. May 7, 2008 #2
    Please recheck your vf, because I find something different using your equation. Then, you can try using Hooke's law with conservation of energy (the spring force is a conservative force). Where is the energy from the moving block and dart going when they slow down to a stop?
  4. May 7, 2008 #3
    I don't think you have the right value for v_f -

    (0.2 kg)*(0 m/s) + (0.05 kg)*(0.1 m/s) = (0.25 kg)*v_f
    v_f = 0.02 m/s

    Although the collision between the dart and the block is inelastic, the compression of a spring does conserve the sum of the kinetic energy and the elastic potential energy!
  5. May 7, 2008 #4
    Thank you I got it. I made a mistake with the calculations b4..but this is the way right?

    Since F= PV => F = (mv)v ==> F= 0.25

    F= kx
    0.25/25 N/m = x
    x= 0.01 m
  6. May 7, 2008 #5
    Why do you say F = pv?
  7. May 7, 2008 #6
    F is not equal to the product of momentum and velocity.... (you can see this by comparing the units of F (kg*m/s^2) and the units of pv (kg*m/s*m/s)).

    It's important to see here that the sum of elastic potential energy and kinetic energy is conserved. So what is the elastic potential energy of a spring compressed distance x?
  8. May 7, 2008 #7
    Sorry I meant P = Fv therefore F = P/V and since P=mv, F = mv/v = m ?
  9. May 7, 2008 #8
    Whoa, you've just said F=mv/v = m.... But force isn't equal to mass!
  10. May 7, 2008 #9
    So 1/2 mv^2 = 1/2 kx^2
    x = 0.01 m
  11. May 7, 2008 #10
    Can you show your substitution? There seems to be a mistake there.
  12. May 8, 2008 #11
    1/2 (0.25 kg)(0.02 m/s) = 1/2 (25 N/m)(x^2)
    x^2 = 0.002 m

    I put the answer for the rong question before..sorry abt that :$
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