Friction / inclined plane question

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SUMMARY

The discussion focuses on calculating the coefficient of kinetic friction for a child sliding down a 28-degree inclined plane, where the final speed is half of what it would be on a frictionless slide. The user employs a free body diagram (FBD) to analyze the forces involved, including gravitational components and frictional force. Key equations include the relationship between gravitational force components and the normal force, leading to the equation: m*g*sin(28) - μk*(m*g*cos(28)) = m*acceleration(x). The problem emphasizes the work done by friction as the child traverses the slope.

PREREQUISITES
  • Understanding of free body diagrams (FBD)
  • Knowledge of gravitational force components
  • Familiarity with the concept of kinetic friction
  • Basic algebra for solving equations
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  • Calculate the coefficient of kinetic friction using the derived equations
  • Explore the work-energy principle in the context of inclined planes
  • Study the effects of varying incline angles on friction and acceleration
  • Investigate real-world applications of friction on inclined surfaces
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I am having trouble setting up the following problem:

A child slides down a slide with a 28 degree incline, and at the bottom her speed is precisely half what it would have been if the slide had been frictionless. Calculate the coefficient of kinetic friction between the slide and the child.

I drew an FBD. I split gravity up into it's X and Y components. I know that Force friction = Mu k * Force normal. Fg(x) = m*g*sin(28). Fg(y) = m*g*cos(28)

Force normal - m*g*cos(28) = m*acceleration(x) = 0

m*g*sin(28) - Mu k * ( m*g*cos(28) ) = m*acceleration(x)

Any ideas?
 
Last edited:
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HINT: How much work does the frictional force do on the slide as it traverses the slope?
 
http://photo-origin.tickle.com/image/69/7/0/O/69708125O733857775.jpg
 
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