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Spring constant

  1. Mar 30, 2010 #1
    1. The problem statement, all variables and given/known data

    A 150g ball is thrown vertically upward from a window 10m above the ground at a speed of 30m/s. At the top of its trajectory, it compresses a spring a distance, y. (spring constant k = 1.5 N/m). If the spring is 45m above the initial starting position of the ball...

    A) how high does the ball go above the ground?
    B) what is the speed of the ball when it strikes the ground?

    2. Relevant equations

    what i am thinking??? use -2g(y-yo) = vf^2-vi^2 then w=kf-ki then w= -1/2 kx^2

    3. The attempt at a solution

    those equations i used for part one...i am not sure where to go with part b
  2. jcsd
  3. Mar 30, 2010 #2


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    This is a prime example of conservation of mechanical energy. Say with an equation that the mechanical energy of the ball right after it is launched is the same as the mechanical energy just before it hits the ground. The spring is irrelevant for part (b).
  4. Mar 30, 2010 #3
    Well the spring should compress the ball so that all the kinetic energy is converted to potential energy.
    Kinetics Energy = (1/2)mv^2
    Potential Energy = mgh

    Do an energy balance.
    (1/2)mv^2 - mg(h+x) = (1/2)*kx^2

    height = 45 meters
    gravity = 9.81 meters/seconds^2
    mass = 150 grams = 0.15 kg
    velocity = 30 m/s

    67.5 J - 1.4715(45+x) = (1/2)*1.5*x^2
    67.5 J - 66.218 J -1.4715x = .75 x^2
    1.282 -.75x^2 -1.4715x = 0
    solve using quadratic equation -> x = 0.653

    b) Assuming no air resistance or heat loss the velocity downwards should be the same as upwards.
    since acceleration is 9.81 m/s^2 we use simple kinematics:

    vf^2 = v0^2 + 2AD

    A = acceleration = 9.81 m/s^2
    D = distance = 10 meters
    v0 = 30 m/s
    vf = you can solve for this. Just simple plug and chug
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