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Homework Help: 3 Questions in regards to WORK AND ENERGY

  1. Oct 22, 2006 #1
    I am attempting 3 questions I have worked them out but to no avail. I wonder if I am on the right track:

    1st Question:

    Starting from rest at the top, a child slides down the water slide at a swimming pool and enters the water at a final speed of 5.00 m/s. At what final speed would the child enter the water if the water slide were twice as high? Ignore friction and resistance from the air and the water lubricating the slide.

    Here I am using the principal of conservation of mechanical energy that can be rewritten as follows:

    Vf = Sqrt(Vo^2 + 2g(ho - hf)

    So I assign a random number (I used 5) to hf and I get:

    5.0 m/s = Sqrt(0 + 19.6(Ho - 5))

    5 = Sqrt(19.6ho - 98)

    25 = 19.6ho - 98

    ho = 6.2755

    So since the H is twice what it would be I used 12.551

    And get:

    Vf = Sqrt(Vo^2 + 19.6(12.551 - 5))

    Vf = 12.16 <------Wrong....is there somewhere I went wrong???


    A 915-kg car starts from rest at the bottom of a drive way and has a speed of 3.00 m/s at a point where the drive way has risen a vertical height of 0.600 m. Friction and the drive force produced by the engine are the only two nonconservative forces present. Friction does -2870 J of work. How much work does the engine do?

    So I used the Work Energy theorum:

    Wnc = 1/2m(Vf^2 - Vo^2) - mg(ho - hf)

    Wnc = 457.5(3)^2 - 8967(-6)

    Wnc = 57919.5

    Do you see anything wrong with my math?


    The power needed to accelerate a projectile from rest to its launch speed v in a time t is 43.0 W. How much power is needed to accelerate the same projectile from rest to a launch speed of 2 v in a time of t?

    I am confused by the relationship here....are there any formulas that would apply here?

    Thanks SO much!

  2. jcsd
  3. Oct 22, 2006 #2
    For 1. Rearrange Vf to get an expression for ho, you know hf = 0. Then just stick 2*ho back into your original expression to find Vf.
  4. Oct 22, 2006 #3
    Do I use the same equation that I used in the beginning?
    Last edited: Oct 22, 2006
  5. Oct 23, 2006 #4
    Does anyone have any ideas for the 2nd and 3rd one?

  6. Oct 23, 2006 #5


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    Staff Emeritus
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    Gold Member

    For question two, do not forget that the engine is doing work not only against gravity, but also against friction.

    For question three, remember that power is the rate at which work is done.
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