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Potential Energy and Conservation of Energy Problems

  1. Jan 8, 2009 #1
    Hi everybody! So this is my first time using this forum. I'm currently taking AP Physics C: Mechanics in high school and I have some questions concerning Potential Energy and Conservation of Energy. Thanks for taking the time to help me!

    1. The problem statement, all variables and given/known data

    32. A 4.0-lb block slides along a horizontal frictionless surface at 8 ft/s. It is brought to rest by compressing a very long spring of spring constant (1/8) lb/ft. The maximum spring compressing is:
    A. 4 ft
    B. 8 ft (Correct Answer)
    C. 16 ft
    D. 2 ft
    E. 45 ft

    33. A block of mass m is initially moving to the right on a horizontal frictionless surface at a speed v. It then compresses a spring of spring constant k. At the instant when the kinetic energy of the block is equal to the potential energy of the spring, the spring is compressed a distance of:
    A. v sqrt(m/2k) (Correct Answer)
    B. (1/2)mv^2
    C. (1/4)mv^2
    D. mv^2/4k
    E. (1/4)sqrt(mv/k)

    34. A 200-lb man jumps out of a window into a fire net 30 ft below. The net stretches 6 ft before bringing the man to rest and tossing him back into the air. The maximum potential energy in (ft*lb) of the net is:
    A. 7200 (Correct Answer)
    B. 6000
    C. 1800
    D. 1200
    E. 600

    2. Relevant equations

    Emech = K + U
    U = 1/2k(x)^2
    U = mgh
    K = 1/2m(v)^2

    3. The attempt at a solution

    32. 1/2 (1/8) x^2 = 1/4 (4) 8^2 <<<< Doesn't work, don't know how I should be setting up the equation.

    33. 1/2 kx^2 = 1/2 mv^2
    kx^2 = mv^2
    x^2 = mv^2/k
    x = v sqrt(m/k) <<<< Why am I missing the 2k?

    34. mgh = 1/2kx^2
    200*32*3 = 1/2*k*6^2
    192000 = 18k
    k = 10,666.67
    U = 1/2 (10,666.67) 6^2 <<<< WAYYYY off.
  2. jcsd
  3. Jan 8, 2009 #2


    User Avatar
    Homework Helper

    Welcome to PF.

    What are you stuck on?
  4. Jan 8, 2009 #3
    After looking over the problem, I realized that i forgot to divide the lb by gravity. That takes care of 32 and 34. For 33, I don't really understand why the answer is vsqrt(m/2k) instead of vsqrt(m/k)
  5. Jan 8, 2009 #4
    Your energy equation is incorrect. If the initial velocity of the block is v, when the block hits the spring, it will lose kinetic energy which will be converted to potential energy. When the potential energy of the spring is equal to the kinetic energy of the block, let the velocity of the block be v1.

    The potential energy gained by the spring is equal to the difference in kinetic energy of the block (initial-current). The current kinetic energy of the block is equal to the potential energy which gives you your answer. Try to set it up.
  6. Jan 8, 2009 #5
    so Usp = Etotal - Ek?
  7. Jan 8, 2009 #6
    The point to realize is that your use of the equation, "1/2 kx^2 = 1/2 mv^2" transfers all energy from the block, to the spring; but, the problem statement requests that you determine the position where Kblock = Uspring. Because Uspring started at zero, the point where the magnitudes are equivalent is 1/2Kblock initial, i.e. the block transfers half of its energy to the spring.
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