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Energy used by a CD player, and height in mgh

  1. Feb 12, 2007 #1
    1. The problem statement, all variables and given/known data

    A portable CD player uses a current of 7.1 mA at a potential difference of 3.5 V. In 65 sec, the CD player uses 1.5925 J of power. Suppose the player has a mass of 0.61 kg. For what length of time could the player operate on the energy required to lift it through a height of 1 m?

    2. Relevant equations

    I thought at first of using 1/2 mv^2 +mgy = 1/2 mv^2 +mgy with initial and final values, but then I realized that I would not be able to solve for v. I am not quite sure how to incorporate the information that they give in order to find out the answer.
     
  2. jcsd
  3. Feb 12, 2007 #2

    ranger

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    Solve for v? I dont follow you. The easiest way to do it, would be to find the power it uses in one second. Then find the total potential energy for 1m. Do you follow me? btw, power is measured in watts; watts = joules/second.
     
  4. Feb 12, 2007 #3
    Wow...that's so much easier! So I would use U=mgh and then set that equal to how much power the CD player uses (which is given) over seconds and solve for how many seconds it takes? mgh = 1.5925 J/ x sec.
     
  5. Feb 12, 2007 #4

    ranger

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    Reread what you just wrote. You're saying that mgh = 1.5925 J. We know the total energy used in 65s is 1.5925J. Find how much energy is needed per second. Then find the amount of PE (mgh). This will be the total amount of energy that will be made available to the player. Since you know how much energy is needed per second, you can find how long it will be powered by mgh.
     
  6. Feb 12, 2007 #5
    So...the energy that is needed per second can be calculated by taking the total energy and dividing it by the time that was originially given. and then mgh can be calculated from the information given...but then how do I solve for the energy needed to move the particle to that distance? Do I multiply my two numbers together?
     
  7. Feb 12, 2007 #6

    ranger

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    Yup
    Its as simple as that.
    Move particle? You lost me with that. I thought we were listening to a CD player.

    Knowing the total PE and the energy the CD player uses per second, cant you figure out how long the CD player will run from the calculated PE?
     
    Last edited: Feb 12, 2007
  8. Feb 12, 2007 #7
    Sorry, it is a CD player...I meant that if I wanted to find out how much time it would take for the CD player to run on that energy, would I have to multiply the two numbers together ?
     
  9. Feb 12, 2007 #8

    ranger

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    This is what I wrote:
    There is no multiplication here. You know the energy it uses per second, and you have the total quantity of energy. Just do the math...
     
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