A man has a total mass m = 83 kg (including skis). He slides directly down a slo

Question

A man has a total mass m = 83 kg (including skis). He slides directly down a slope with an angle = 24° as shown with a constant speed of 73 kph in a straight line. The coefficient of kinetic friction is 0.28.

1.What is the rate at which gravity is doing work on him? (use the total mass).

2.Now determine the rate at which friction is doing work on him.

3.What is the rate at which air resistance is doing work on him?

4.What is the rate at which all the forces acting on him are doing work on the skier?

Explanation / Answer

here,

1 kph = 0.278 m/s

mass of man, m = 83 kg
Angle of slope, A = 24 degrees

velocity, v = 73 kph = 20.278 m/s

coefficient of kinetic fgriction, uk = 0.28

Part A:

Since power = Work done /time = Force * distance/time

p = Force*velocity -----------------------(1)

pg = mg*SinA*v ( Pg power duer to gravity)
pg = 83*9.81*Sin24*20.278
Pg = 6715.611 Watts

Part B:
Fron newton Second law, Fnet = 0
Force - Fritctional force = 0
F = Ff
F = uk*mg*Cos24
F = 0.28*83*9.81*Cos24
F = 208.274 N

Power due to friction = F*v
Pf = 208.274*20.278
Pf = 4223.38 watts

Part C:
Force Of air resistance, Fr = Fg - Ff ( Fg is force duue to gravity)
Fr = mg*Sin24 - uk*mg*Cos24
Fr = 83*9.81*Sin24 -0.28*83*9.81*Cos24
Fr = 122.903 N

Therefore Power due to resistance, Pr
Pr = 122.903 * 20.278
Pr = 2492.227 Watts

Part D:
Net Power = Pg - (Pf + Pr)
P = 6715.611 -(4223.38 + 2492.227)
P = 0.004 Watts

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