Question

1. at some fancier sledding hills there is a motorized rope loop used to pull sledders back to the top of the hill. the hillside is inclided at 25 degrrees and the rope pulls sledders at a constant 2.5 m/s. sleds arent quite a frictionless a good skis, so let's say that thete is a coefficient of kinetic friction of =0.05 between the sled and the snow.

a) how much power must the drive motor provide to pull a single sledder and sled with a combined mass of 180 kg up the hill under these conditions

b) if the sled run is 40m long, how much work does the motor do on the sledder and sled during the trip back to the top of the hill?

Answer 1

a)
Gravitational force along the incline, Fg = mg * sin
Where m is the combined mass of the sledder and the sled, m = 180 kg, = 25 degrees.
Fg = 180 * 9.8 * sin(25)
= 745.5 N

Frictional force along the incline, Ff = k * mg * cos
Where k is the coefficient of kinetic friction,
Ff = 0.05 * 180 * 9.8 * cos(25)
= 79.94 N

Total force on the mass, F = Fg + Ff
= 745.5 + 79.94
= 825.4 N

Power delivered = Force * velocity
= 825.4 * 2.5
= 2063.6 W

b)
Work done against gravity, Wg = mg * d * sin
Where d = 40 m
Wg = 180 * 9.8 * 40 * sin(25)
= 29819.94 J

Work done against friction, Wf = Ff * d
= 79.74 * 40
= 3197.5 J

Total work done, W = Wg + Wf
= 29819.94 + 3197.5
= 33017.4 J