Asked by Amy
A 60.0-kg skier with an initial speed of 12.0 m/s coasts up a 2.50-mhigh rise and angle is 35 degrees.Find her final speed at the top, given that the coefficient of friction between her skis and the snow is 0.0800. (Hint: Find the distance traveled up the incline assuming a straight-line path as shown in the figure.)
Answers
Answered by
Henry
Wt. = Fg = m*g = 60kg * 9.8N/kg=588 N.=
Wt. of skier.
Fp=588*sin35 = 337 N.=Force parallel to
incline.
Fv = 588*cos35 = 482 N. = Force perpendicular to incline.
Fk = u*Fv = 0.08 * 482 = 38.5 N. = Force
of kinetic friction.
d =h/sinA = 2.5/sin35 = 4.36 m.
Ek + Ep = Ekmax - Fk*d
Ek = Ekmax-Ep-Fk*d
Ek=0.5*60*12^2-588*2.5-38.5*4.36=2682 J.
Ek = 0.5m*V^2 = 2682 J.
30*V^2 = 2682
V^2 = 89.4
V = 9.5 m/s = Final velocity.
Wt. of skier.
Fp=588*sin35 = 337 N.=Force parallel to
incline.
Fv = 588*cos35 = 482 N. = Force perpendicular to incline.
Fk = u*Fv = 0.08 * 482 = 38.5 N. = Force
of kinetic friction.
d =h/sinA = 2.5/sin35 = 4.36 m.
Ek + Ep = Ekmax - Fk*d
Ek = Ekmax-Ep-Fk*d
Ek=0.5*60*12^2-588*2.5-38.5*4.36=2682 J.
Ek = 0.5m*V^2 = 2682 J.
30*V^2 = 2682
V^2 = 89.4
V = 9.5 m/s = Final velocity.
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