Kinetic Friction and skis thrust

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SUMMARY

The discussion centers on calculating the forces acting on Sam, who weighs 71.0 kg and descends a 51.0 m high slope at an angle of 11.0 degrees using jet-powered skis with a thrust of 210 N. The participant correctly identifies the gravitational force components, calculating the gravitational force (FG) as 695.8 N, the normal force (FN) as 683 N, and the downhill gravitational component as 133 N. The total force acting down the slope is 343 N, combining the thrust and the gravitational component. The participant seeks clarification on whether the problem requires finding the coefficient of friction.

PREREQUISITES
  • Understanding of Newton's second law (Net Force = ma)
  • Knowledge of kinematic equations
  • Familiarity with free body diagrams
  • Basic principles of kinetic friction
NEXT STEPS
  • Calculate the coefficient of kinetic friction using the formula: friction force = coefficient * normal force.
  • Explore the implications of thrust force on motion dynamics in inclined planes.
  • Review the concepts of gravitational force components on slopes.
  • Study the effects of varying angles on friction and thrust in physics problems.
USEFUL FOR

Students studying physics, particularly those focusing on mechanics involving forces on inclined planes, as well as educators looking for practical examples of applying Newton's laws and friction concepts.

BlakeGriffin
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Homework Statement


Sam, whose mass is 71.0 kg , takes off down a 51.0 m high, 11.0 degree slope on his jet-powered skis. The skis have a thrust of 210 N . Sam's speed at the bottom is 41.0 m/s .


Homework Equations


I'm not sure which equation(s) to use.
Net Force= ma and some kinematic equations maybe?


The Attempt at a Solution


This is the free body diagram that I came up with.
http://img291.imageshack.us/i/fbdy.jpg/ (It's not to scale)
There is also the thrust force coming from the skis in the opposite direction of the friction force. I couldn't fit them all in. Btw the dashed lines are the components of the gravity force.

If my free body diagram is right, then normal force = FG Cos(theta) which is one of the components of the gravity force. The other component of gravity is FG Sin(theta)

FG is mg = (71 KG)(9.8 m/s^2)=695.8 N
FN=FG Cos(theta)=695.8(cos(11))0=683N
The other component of FG is =FGSin(theta)=695.8(Sin(11))=133N

Since this component of gravity and the thrust force coming from the skis both point down the hill (or whatever) I get 133+210=343N pointing in that direction.

I don't know where to go from here.. Can anyone help me?
 
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Ummm...what's the problem asking? Is it looking for the coefficient of friction or something?
 

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