Question
A 61-kg skier, coasting down a hill that is an angle of 23 to the horizontal, experiences a force of kinetic friction of magnitude 72N. The skier's speed is 3.5m/s near the top of the slope. Determine the speed after the skier has travelled 62m downhill. Air resistance is negligible.
Answers
Initial Ke = (1/2) m v^2 = (1/2)(61)(3.5)^2
drop in potential energy = m g h
I assume that the 62 m is along the slope so
h = 62 sin 23 which is drop down toward center of earth
so drop in potential = 61*9.81*62 sin 23
work done against friction = friction force * distance = 72*62
so
(1/2)(61)v^2 = (1/2)(61)(3.5)^2+ 61*9.81*62 sin 23 - 72*62
solve for v
drop in potential energy = m g h
I assume that the 62 m is along the slope so
h = 62 sin 23 which is drop down toward center of earth
so drop in potential = 61*9.81*62 sin 23
work done against friction = friction force * distance = 72*62
so
(1/2)(61)v^2 = (1/2)(61)(3.5)^2+ 61*9.81*62 sin 23 - 72*62
solve for v
Fp = mg*sin23 = 597.8*2in23=233.6 N.=
Force parallel to the hill.
Fv = mg*cos23 = 597.8*cos23=550.3 N.=
Force perpendicular to the hill.
Fk = 72 N.
a = (Fp-Fk)/m = (233.6-72)/61=2.65 m/s^2
V^2 = Vo^2 + 2a*d
V^2 = (3.5)^2 + 2*2.65*62 = 340.85
V = 18.5 m/s.
Force parallel to the hill.
Fv = mg*cos23 = 597.8*cos23=550.3 N.=
Force perpendicular to the hill.
Fk = 72 N.
a = (Fp-Fk)/m = (233.6-72)/61=2.65 m/s^2
V^2 = Vo^2 + 2a*d
V^2 = (3.5)^2 + 2*2.65*62 = 340.85
V = 18.5 m/s.