Determine the distance the sled will slide before coming to rest.

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SUMMARY

The discussion focuses on calculating the distance a sled will slide down a hill inclined at 6 degrees with an initial speed of 12 m/s and a coefficient of friction of 0.14. The key equations involved include Newton's second law (F=ma) and the kinematic equation vFsquared=vIsquared + 2a(d). The participants emphasize the importance of drawing a free body diagram and correctly identifying forces, including gravitational force and frictional force, to determine acceleration and ultimately the distance slid before coming to rest.

PREREQUISITES
  • Understanding of Newton's second law (F=ma)
  • Familiarity with kinematic equations
  • Knowledge of friction and the coefficient of friction
  • Ability to draw and interpret free body diagrams
NEXT STEPS
  • Study the derivation of the kinematic equation vFsquared=vIsquared + 2a(d)
  • Learn how to calculate forces in free body diagrams
  • Explore the relationship between mass, friction, and acceleration
  • Practice problems involving inclined planes and friction
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Students studying physics, particularly those focusing on mechanics and dynamics, as well as educators seeking to enhance their teaching methods in these topics.

ExtendedG
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Homework Statement


A sled takes off from the top of a hill inclined at 6 degrees to the horizontal. The sled's initial speed is 12 m/s. The coefficient of friction between the sled and the snow is 0.14. Determine how far the sled will slide before coming to rest.


Homework Equations


F=ma
vFsquared=vIsquared + 2a(d)
vf= velocity final vI= initial velocity a=accel. d=displacement.


The Attempt at a Solution


Well I didn't get very far because I don't have enough variables, I need mass! I obviously need to find accelerating, what would my f=ma statement be? I thought it would be Fapp - Ffriction = ma...help?
 
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Perhaps worth remembering the definition of coefficient of friction. It's the ratio of two forces one of which depends on the mass. This means that mass sometimes cancels in some problems so it's not allways needed.

Draw the free body diagram and write equations for the forces.
 
CWatters said:
Perhaps worth remembering the definition of coefficient of friction. It's the ratio of two forces one of which depends on the mass. This means that mass sometimes cancels in some problems so it's not allways needed.

Draw the free body diagram and write equations for the forces.
F=ma
Fg - Ff = ma
mgsin(theta) - umgcos(theta) = ma
I have two masses to remove, not just one :/ So i can't just simply simplify to:
gsin(theta) - ugcos(theta) = a right?
 

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