How Does the Kinetic Energy Theorem Apply to Bonny Blair's Skating Problem?

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Homework Help Overview

The problem involves applying the kinetic energy theorem to a scenario where Bonny Blair skates a certain distance at varying speeds, and the focus is on determining the work done by friction during her deceleration.

Discussion Character

  • Exploratory, Conceptual clarification, Mathematical reasoning

Approaches and Questions Raised

  • Participants discuss the kinetic energy theorem and its application to the problem, with one suggesting that the change in kinetic energy corresponds to the work done by friction. Others raise questions about the specific calculations needed to determine the work done.

Discussion Status

The discussion is ongoing, with participants exploring different aspects of the kinetic energy theorem and its implications for the problem. Some guidance has been offered regarding the relationship between kinetic energy and work, but no consensus has been reached on the specific calculations or methods to be used.

Contextual Notes

There are mentions of potential methods and considerations regarding the distance over which friction acts, but no definitive information is provided about the values or assumptions that may be influencing the calculations.

Femme06Fatale
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Bonny Blair skated 5.00 x 10^2 m with an average speed of 12.92 m/s. Suppose she corssed the finish line at this speed and then skated freely until her speed was 8.00 m/s. If her was mwas 55.0 kg, how much work was done by friction?

Can anybody explain the kinetic energy theorem or help me work through this problem?
 
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The kinetic energy theorem says her variation in kinetic energy is equal to the work done on her.

If you can find by how many Joules her kinetic energy varied, then this same number is equal to the work done on her:

[tex]\Delta K = W[/tex]
 
[tex]W_{friction}=F_{friction}d\cos{\theta}[/tex]

[tex]F_{net}=ma[/tex]

Can you figure it out using those (careful what you use for [itex]d[/itex])?
 
Oh, yes, quasar's method may be easier.
 

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