Inclined Plane (friction)

  • Thread starter eestep
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  • #1
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Homework Statement


An object of mass 13 kg is placed on incline with friction. Incline is originally horizontal and then raised slowly and at 22 degrees, mass begins to slide down incline. It is found that coefficient of kinetic friction is .83 times coefficient of static friction. If incline is raised to 41.3 degrees and object enters bottom of ramp at some velocity and its velocity at top of incline is 24 m/s, what would be its velocity at bottom of incline in m/s if height (not length) of ramp is 10.5 meters? Answer is 29.61.
m=13 kg
[tex]\theta[/tex]=22
[tex]\mu[/tex]k=.83[tex]\mu[/tex]s
[tex]\phi[/tex]=41.3
v0=24 m/s



Homework Equations


vf2=v02+2a[tex]\Delta[/tex]x



The Attempt at a Solution


N=118.24
 
Last edited:

Answers and Replies

  • #2
rl.bhat
Homework Helper
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When you keep an object on the inclined plane and just start moving,
the normal reaction N is m*g*cosθ
the frictional force f is m*g*sinθ.
Hence μs = f/N = tanθ.
Find μk.
Net acceleration along the inclined plane is a = g(sinθ + μk*cosθ)
Find the length of the inclined plane. sinθ = h/L.
Use the relevant equation and find vf.
 
Last edited:
  • #3
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Angle changes, and I am confused as far as when each one is appropriate. I appreciate your assistance but am not getting the answer by just plugging in 22 degrees.
 
  • #4
rl.bhat
Homework Helper
4,433
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Since the object is going form bottom to top, the net acceleration is

a = g(sinθ + μk*cosθ). Here θ = 41.3 degrees.

vf^2 = vi^2 - 2*g*(sinθ + μk*cosθ)*L.

vf is given. Find vi.
 
  • #5
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I finally got the answer. Thanks!
 

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