Question

Suppose the ski patrol lowers a rescue sled and victim, having a total mass of 95.0 kg, down a θ = 50.0° slope at constant speed. The coefficient of friction between the sled and the snow is 0.100.

(a) How much work is done by friction as the sled moves 30.0 m along the hill?


(b) How much work is done by the rope on the sled in this distance?


(c) What is the work done by gravity on the sled?


(d) What is the total work done?

Answer 1

M*g = 95 * 9.8 = 931 N. = Wt. of load.

Fp = 931*sin50 = 713.2 N. = Force parallel to the incline.

Fn = 931*Cos50 = 598.4 N. = Normal = Force perpendicular to the incline.

Fk = u*Fn = 0.10 * 598.4 = 59.84 N. =
Force of kinetic friction.

a. Work = Fk*d = 59.84 * 30 = 1795 N.

b. Work = Fp*d =

c. Work = Mg*d

d. Sum of all work.

Answer 2

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