A skier of mass 80.0 kg slides down a slope inclined at an angle of 20.0 with th
ID: 1964167 • Letter: A
Question
A skier of mass 80.0 kg slides down a slope inclined at an angle of 20.0 with the horizontal. Theskier starts from rest and travels a distance of 40.0 m. (Note: This is the distance traveled along the slope,not just the vertical distance.) The nal speed of the skier is 12.0 m/s. Ignore air resistance. (a) What is thenal kinetic energy of the skier? (b) What is the total work done on the skier? (c) What is the work done bygravity? (d) What is the work done by the normal force? (e) What is the work done by friction? (f) What isthe force of friction (assuming it is constant)? (g) What is the coefcient of kinetic friction?
Explanation / Answer
mass = m = 80 kg
angle = = 20 degrees
initial speed = u = 0
final speed = v = 12 m/s
Distance = L= 40 meters
Vertical distance travelled =h = Lsin(20) = 40sin(20) = 13.68 meters
Work by friction =Wf
a. Final Kinetic Energy =0.5mv^2 = 0.5 x 80 x 1262 = 5760 Joules
b. Work done on Skier = Change in Kinetic Energy = 5760 - 0 = 5760 Joules
c. Work done by gravity = mgh = 80 x 9.8 x 13.68 = 10725.12 Joules
d. Work done by normal force = 0 ( as it is always at 90 degrees to displacement)
e.
Now,
from Work energy theorem,,
0.5mv^2 - 0 = mgh + Wf
0.5 x 80 x 12^2 - 0 = 80 x 9.8 x 13.68 + Wf
5760 = 10725.12 + Wf
Wf = -4965.12 Joules
WOrk done by frciton = -4965.12 Joules
f. Force of Friction f = Wf/L = 4965.12/40 = 124.128 Newtons
g. We have,
f = N
124.128 = x 80 x 9.8 cos(20)
Coefficient of frciton = = 0.1685
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