A skier traveling 12.4 m/s reaches the foot of a steady upward 17.4° incline and glides 13.9 m up along this slope before coming to rest. What was the average coefficient of friction?
How would solve this?
2006-11-02 06:51:16 · 3 answers · asked by Isabel G 1 in Science & Mathematics ➔ Physics
The work done by friction is the difference between the kinetic energy of the skier at the bottom of the slope and the potential energy of the skier when he stops. So KE (initial) = PE (final) + W. KE is 0.5*m*v^2 and PE is mgh, where h is the height at rest. W is Fd = umgd*cos(theta), where u is the coefficient of friction, d is the path length traveled, and theta is the angle of inclination. The height h will be d*sin(theta). The equation becomes 0.5mv^2 = mgd*sin(theta) + umgd*cos(theta). The mass is constant and unknown, so we can factor it out, leaving 0.5v^2 = gd*sin(theta) + ugd*cos(theta). u is your only unknown, so you can solve algebraically.
2006-11-02 06:59:12 · answer #1 · answered by DavidK93 7 · 1⤊ 0⤋
You know the initial kinetic energy = 12.4^2/2 (m) = 76.9m
The final potential energy = mgh = mg(13.9sin17.4) = 40.9m
So the diff of 36m is due to engergy lost from friction.
Frictional energy = 36m = Fr (distance) = Fr (13.9)
The Fr = uN = umgcos17.4
So now we have 36m = umgcos17.4, the m's cancel and you
can find u, the coefficient of friction.
2006-11-02 15:02:02 · answer #2 · answered by Jim C 3 · 0⤊ 0⤋
The Kinetic Energy, KE, at the foot of the incline is converted to Work plus Potential Energy, PE.
Given: velocity at foot of incline, v=12.4m/s
length of incline=13 .9m
angle of incline=17.4degrees.
Weight of skier=mg
Component of his weight normal to incline=
mg*cos17.4.
Component of his weight along incline=
mg*sin17.4
Force of friction, f=uR where R=the force normal to the
incline=mgcos17.4
Therefore,
f=u*mgcos17.4
Net force along the incline, F=force of friction-component of his weight along incline.
F=u*mgcos17.4-mgsin17.4
Work=F*s where s is the displacement=13.9
=(u*mgcos17.4-mgsin17.4)*13.9
KE=1/2mv^2
=1/2m*12.4^2
PE=Weight of skier*vertical height
=mg*13.9sin17.4
Based on conservation of energy principle:
KE=PE+Work
Substitute known values:
1/2m*12.4^2=mg*13.9sin17.4+
(u*mgcos17.4-mg*sin17.4)*13.9
m cancels out.
76.9=g*13.9*(.299+u*.954-.299)
76.9=g*13.9*u*.954
76.9=130u
=0.59
2006-11-04 08:34:41 · answer #3 · answered by tul b 3 · 0⤊ 0⤋