Energy used by a CD player, and height in mgh

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SUMMARY

The discussion focuses on calculating the operational time of a portable CD player using its energy consumption and potential energy. The CD player operates at a current of 7.1 mA and a voltage of 3.5 V, consuming 1.5925 J over 65 seconds. To determine how long the player can run on the energy required to lift it through a height of 1 meter, participants suggest using the formula U=mgh to equate potential energy with the power consumption of the CD player. The key takeaway is that by calculating the energy used per second, one can find the duration the CD player will operate based on the potential energy available.

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  • Understanding of electrical power calculations (Watts = Joules/second)
  • Familiarity with gravitational potential energy formula (U=mgh)
  • Basic knowledge of energy conservation principles
  • Ability to perform unit conversions and algebraic manipulations
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  • Explore energy consumption metrics for portable electronic devices
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Homework Statement



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?

Homework 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.
 
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Solve for v? I don't 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.
 
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.
 
map7s said:
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.

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.
 
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?
 
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.
Yup
and then mgh can be calculated from the information given
Its as simple as that.
but then how do I solve for the energy needed to move the particle to that distance? Do I multiply my two numbers together?
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, can't you figure out how long the CD player will run from the calculated PE?
 
Last edited:
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 ?
 
This is what I wrote:
Knowing the total PE and the energy the CD player uses per second, can't you figure out how long the CD player will run from the calculated PE?
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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