Average rate of energy transfer & rate of energy dissipation.

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Homework Help Overview

The discussion revolves around calculating the average rate of energy transfer into gravitational potential energy and the rate of energy dissipation for a boy running up stairs and a car braking to rest. The subject area includes concepts of energy transfer, gravitational potential energy, and kinetic energy.

Discussion Character

  • Exploratory, Conceptual clarification, Mathematical reasoning

Approaches and Questions Raised

  • The original poster expresses uncertainty about how to calculate the required rates of energy transfer and dissipation. Participants suggest relevant equations and clarify the need for time in the calculations. There is discussion about the applicability of the formula E=mgh for different scenarios.

Discussion Status

Participants are actively engaging with the problem, offering guidance on relevant equations and clarifying the distinction between the two scenarios presented. There is an exploration of different formulas, but no consensus has been reached regarding the final calculations.

Contextual Notes

The original poster has not provided all necessary details for the calculations, and there is an emphasis on understanding the distinction between energy transfer in vertical motion versus energy dissipation in horizontal motion.

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Homework Statement


1.) A 70kg boy runs as fast as he can up a flight of stairs which has vertical height of 10metres. He achieves this in 8 seconds. Calculate the average rate of energy transfer into gravitational potential energy that must have occurred.

2.) A700kg car traveling at 20ms-1 breaks to rest in 5 seconds. Calculate the rate of energy dissipation the car must achieve.

Hello, i am not certain how to calculate these.
 
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Hello,

i thought maybe mgh and mgh/t?

70x9.8x10?

700x9.8x(20)/5
 
Last edited:
E=mgh only works for vertical changes. You will use it in the first one, but not in the second one.

If I understand the question, you almost have the first one. However, it asks for the rate, which implies a time. So, you have to put the answer in the form J/s.

The second one requires a different formula, one which uses kinetic energy. See if you can find this formula; there's a handy little sticky with some of the more common formulas at the top of this page.
 
Vidatu said:
E=mgh only works for vertical changes. You will use it in the first one, but not in the second one.

If I understand the question, you almost have the first one. However, it asks for the rate, which implies a time. So, you have to put the answer in the form J/s.

The second one requires a different formula, one which uses kinetic energy. See if you can find this formula; there's a handy little sticky with some of the more common formulas at the top of this page.

Hello,

1/2mv^2/t

?
 

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