Average rate of energy transfer & rate of energy dissipation.

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SUMMARY

The discussion focuses on 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. For the boy, the average rate of energy transfer can be calculated using the formula E=mgh, resulting in 70kg * 9.8m/s² * 10m divided by 8 seconds, yielding a rate in J/s. For the car, the rate of energy dissipation requires the kinetic energy formula, specifically 1/2mv² divided by the time taken to stop, which is 5 seconds.

PREREQUISITES
  • Understanding of gravitational potential energy (E=mgh)
  • Knowledge of kinetic energy (E=1/2mv²)
  • Basic principles of energy transfer and dissipation
  • Familiarity with units of measurement (Joules, seconds)
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  • Research the concept of energy transfer rates in physics
  • Study the derivation and applications of the kinetic energy formula
  • Explore examples of energy dissipation in real-world scenarios
  • Learn about the conservation of energy in mechanical systems
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Students studying physics, particularly those focusing on mechanics and energy concepts, as well as educators looking for practical examples of energy transfer and dissipation calculations.

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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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