A skier has just begun skiing down a 30 degree slope. She is going at a constant speed. What is her coefficient of sliding friction?
(in the previous problem, the coefficient of sliding friction was .20 with an undetermined acceleration rate)
2006-12-13 06:37:22 · 6 answers · asked by The Answer 1 in Science & Mathematics ➔ Physics
Since
F=mg(cos 30 - u sin (30))=0
u=cos(30)/sin(30)= .866/.5=1.732
2006-12-13 06:56:28 · answer #1 · answered by Edward 7 · 0⤊ 2⤋
The weight of the skier can be resolved
along and perpendicular to the inclined plane.
force along the inclined plane
=mgsin30[m=mass,g=accn.due to graviy=9.8m/sec^2]
=m*9.8*1/2
force in perpendicular direction
=mg cos30
=m*9.81*sqrt3/2
=normal reaction
Therefore kinetic frictional force
=k*m*9.8*sqrt3/2
This frictional force opposes the
driving component due to gravity.
If the body is not accelerating these
have got to cancel each other
therefore
k*m*9.8*sqrt3/2
=m*9.8*1/2
k=[sqrt3/2]/1/2
=sqrt3
=1.732
[the value of coefficient is quite large.
Normally it should be less than .4 or .5
2006-12-13 06:55:50 · answer #2 · answered by openpsychy 6 · 0⤊ 1⤋
sin(30)w = (u_k)cos(30)w
(u_k) = 0.57735054
coefficient of sliding friction is 0.57735054
2006-12-13 07:12:37 · answer #3 · answered by Anonymous · 0⤊ 0⤋
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2016-10-14 21:14:18 · answer #4 · answered by ? 4 · 0⤊ 0⤋
a= sin(30)g-mu(cos(30)g)
0= 4.9- mu(8.5)
-4.9= -mu(8.5)
-mu= -.576
mu= 0.576
2006-12-13 07:13:56 · answer #5 · answered by 7 · 0⤊ 1⤋
why do u ask hard questions?
2006-12-13 06:50:25 · answer #6 · answered by Anonymous · 0⤊ 2⤋