Askier is pulled by a tow rope up a frictionless ski slope that makes an angle of 7.1° with the horizontal. the rope moves parallel to the slope with a constant speed of 0.60 m/s. the force of the rope does 600 j of work on the skier as the skier moves a distance of 4.6 m up the incline. (a) if the rope moved with a constant speed of 2.8 m/s, how much work would the force of the rope do on the skier as the skier moved a distance of 4.6 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) 0.60 m/s and (c) 2.8 m/s?

W = 600 J

Explanation:

Let's use the work equation to find the force

     W = f d

     F = W / d

     F = 600 / 4.6

     F = 130.4 N

a) The expression for power is

      P = W / t = f .v

      W = f v t

We need to calculate the rise time, which can be found by kinematics

      v = x / t

      t = x / v

      t = 4.6 /2.8

      t = 1.64 s

We calculate

     W = 130.4 2.8 1.64

     W = 600 J

b)  we repeat the calculations changing the speed  to v = 0.60m/s

      t = 4.6/0.6    

      t = 7.67 s

     W = 130.4 0.6 7.67

     W = 600 J

You can see that work is always the same what changes is the power

the device is called “ rocket engine”

answer; equivalent to a one(1) horsepower motor;