A skier is pulled by a tow rope up a frictionless ski slope that makes an angle

Question

A skier is pulled by a tow rope up a frictionless ski slope that makes an angle of 5.4° with the horizontal. The rope moves parallel to the slope with a constant speed of 1.9 m/s. The force of the rope does 710 J of work on the skier as the skier moves a distance of 6.1 m up the incline. (a) If the rope moved with a constant speed of 2.0 m/s, how much work would the force of the rope do on the skier as the skier moved a distance of 6.1 m up the incline? At what rate is the force of the rope doing work on the skier when the rope moves with a speed of (b) 1.9 m/s and (c) 2.0 m/s?

Explanation / Answer

work = mgh = 710

m*9.81*6.1sin(5.4) = 710

m = 126.07 kg

a)

It is same as before as work done is = mgh which is independert of velocity

b)

F = mgsin(theta) = 126.07*9.81*sin(5.4) = 116.4 N

rate of doing work = Power = F.V = 116.4*1.9 = 221.14 Watts

c)

116.4 * 2 = 232.8 Watts

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