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    Pogo Stick -- Kinetic Energy

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    A child's pogo stick stores energy in a spring with a force constant of 2.50 x 10^4 N/m. At position A (XA= -0.100 m), the spring compression is a maximum and the child is momentarily at rest. At position B (Xb=0) the spring is relaxed and the child is moving upward. At position C the child again is momentarily at rest at the top of the jump. The combined mass of child and pogo stick is 25 kg,
    k = 2.50 x 104 N/m
    xA= -0.100 m
    xB = 0
    mchild+pogo = 25 kg

    a) For this problem it is necessary to find the total energy of the system. According to the conservation of energy theorem all of the energies involved will be the same at each point or letter that is shown in the figure. Therefore, the energy found for A will be the same energy that is for part B and so on for C. However, the spring potential, gravitational potential and kinetic energy will all change to accommodate for the movement of the boy. In this case I am given that the at x=0 the potential and elastic energies are zero, therefore, in order to find the total energy of the system I will find the energy at point A.
    I am able use the equation for the spring below because x is less than 0. When x is greater than zero than the spring equation will be equal to zero.

    EA= KA+ UsA+ mgxA
    EA = 1/2mv2 +1/2kx2 + mgx
    Since the boy is at rest the velocity will be zero Also,
    EA= 1/2kx2+mgx
    EA = ½(2.50 x 104 N/m)(-0.100 m) 2 +(25 kg)(9.8 m/s2)(-0.100 m)
    EA = 125 N*m + -24.5 N*m
    EA = 100.5 N*m'

    b) In order to find xc we can use the energy from part A since we know that the total energy is the same at each position of the 3 diagrams. However, we know that at position C the potential of the spring is zero because now that the spring is not attached to the ground anymore it is greater than zero. And as stated before when x is greater than zero the spring potential is also equal to zero. Therefore:

    EC = KC+ Us + mgxC
    EC = ½mv2 + 1/2kx2 +mgx
    EC = 0 + 0 + mgx
    EC = mgx
    x = EC/mg
    xC = 1005. N*m/(25 kg)(9.8 m/s2)
    xC = 0.410 m

    c) Now if we want to find the speed when x=0 we can use the diagram for position B. At this point since x=0 the gravitational potential and spring potential are both zero. Therefore the total energy is only equal to the kinetic energy of the system at that time.

    EB = 1/2mv2 +1/2kx2 + mgx
    EB = 1/2mv2+ 0 + 0
    EB = 1/2mv2
    v2 = 2(EB)/m
    v = [(2EB)/m]1/2
    v = 2.84 m/s

    d) Determine the value of x for which the kinetic energy of the system is a maximum

    e) Calculate the child's maximum upward speed.

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    Solution Summary

    The solution is provided in an attachment, including two diagrams.