Energy used by a CD player, and height in mgh

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Homework Help Overview

The discussion revolves around a problem involving the energy consumption of a portable CD player, specifically calculating how long it can operate based on the energy required to lift it through a height of 1 meter. The subject area includes concepts from energy, power, and mechanics.

Discussion Character

  • Exploratory, Conceptual clarification, Mathematical reasoning

Approaches and Questions Raised

  • Participants explore the relationship between power, energy, and potential energy, with some suggesting to calculate the power used per second and others discussing the use of the potential energy formula U=mgh. Questions arise about how to relate the total energy used by the CD player to the energy required for lifting.

Discussion Status

The discussion is active, with participants providing guidance on how to approach the problem. There is an exploration of different interpretations regarding the calculations needed to determine the operating time of the CD player based on the potential energy available.

Contextual Notes

Participants are working within the constraints of the problem's given values, including the current, potential difference, and mass of the CD player, while also considering the energy used over a specified time period.

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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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