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Homework Help: Urgent: Totally lost regarding conservation of energy

  1. Mar 23, 2008 #1
    These were review problems for the test I have tomorrow. My teacher was not present the day these were given, so I could not ask him for help. The answers were given for these problems, but in terms of getting to these answers, I am totally lost. Can someone please explain to me how to achieve the answers to these questions (answers are below questions in my explanations)? Thanks for any assistance. If you can only do one, then that's fine. Any explanations will do because I need to study with them for the test tomorrow afternoon. Again, thanks a bunch.

    1. A 0.280 kg block on a vertical spring with a spring constant of 5.00 x 10^3 N/m is pushed downward, compressing the spring 0.110 m. When released, the block leaves the spring and travels upward vertically. How high does it rise above the point of release?


    Wouldn't this have to do with elastic energy, so 1/2KX^2? The answer is 11m, but I don't know what method to use to achieve this answer.


    2. A 80 kg trampoline artist jumps vertically upward from the top of a platform with a speed of 6.0 m/s.

    How fast is he going as he lands on the trampoline, 2 m below?

    If the trampoline behaves like a spring of spring constant 5.2 × 104 N/m, how far does he depress it?

    (Ignore small changes in gravitational potential energy.)
    For the first part, I thought you would do: PE + KE (Before) = PE + KE (After). KE (before) and PE (after) would be 0 because he doesn't have any kinetic energy before he jumps, and he doesn't have any potential energy after he jumps. The answer is 8.67m/s, and again, I don't know what method to use to get the answer.

    For the second sub-question, I thought you would again use the elastic energy equation, but alas I was wrong. The answer for the second question is 0.34m. I just don't know how to get this answer.

    3. A 14 kg child descends a slide 2.50 m high and reaches the bottom with a speed of 2.46 m/s. How much thermal energy due to friction was generated in this process?
    I would assume PE=KE though. The answer is 301 J.

    4. A 68 kg person escapes from a burning building by jumping from a window 25 m above a catching net. Assuming that air resistance exerts a 95 N force on the person during the fall, determine the person's velocity just before hitting the net.
    I would assume PE=KE though. The answer is -20.5 m/s.
    5. A roller coaster, shown in the figure (ha = 50 m, hb = 28 m, hc = 12 m), is pulled up to point 1 where it and its screaming occupants are released from rest. Assuming no friction, calculate the speed at points 2, 3, 4. The roller coaster cart is 500kg.

    http://i26.tinypic.com/hurosi.gif - Link to figure.

    Totally lost on this one. The answers are:

    m/s (at point 2)

    m/s (at point 3)

    m/s (at point 4)

  2. jcsd
  3. Mar 23, 2008 #2


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    Homework Helper

    Assuming no energy losses
    All the elastic potential energy is converted to gravitational potential energy.
    You know that the elastic potential energy=[itex]\frac{1}{2}kx^2[/itex] and that should be the same as the gravitational potential energy. Do you know an equation to find the gravitational potential energy of an object of mass,m, at a height,h, in a gravitational of field of field strength,g?

    All the gravitational pe is converted into energy to cause the child to move at 2.46m/s and energy in overcoming friction.
    Same concept for Q4

    For Q5. If you are ignoring frictional effects. From point 1 to point 2, there is a change in height so the gravitational pe would change. What do you think all this energy is converted into?
    Last edited: Mar 23, 2008
  4. Mar 23, 2008 #3


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    Staff Emeritus
    Science Advisor

    There are three energies involved here - 1. spring mechanical (potential) energy, 2. gravitational potential energy, and 3. kinetic energy.

    When released, the spring energy (1/2 kx2, where x = 0.110 m) is changes the gravitational potential energy (mgh) and kinetic energy of the block over the 0.110 m. So equate those. Then from the KE of the mass (1/2 mv2), that gives the initial velocity for the second part of the problem, which is a vertical launch.

    See this for reference - http://hyperphysics.phy-astr.gsu.edu/hbase/traj.html#tra1

    for work done by (or lost to friction), one applied the definition of work = force * distance.
  5. Mar 23, 2008 #4
    Astronuc; if we do not know the height nor the velocity, how do we solve the equation?

    Is it 1/2kx^2+mgh=1/2mv^2 for problem 1, part one?
  6. Mar 24, 2008 #5
    Velocity at top most point will be zero:smile:
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