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 7.0° with the horizontal. The rope moves parallel to the slope with a constant speed of 1.4 m/s. The force of the rope does 750 J of work on the skier as the skier moves a distance of 3.5 m up the incline. (a) If the rope moved with a constant speed of 2.1 m/s, how much work would the force of the rope do on the skier as the skier moved a distance of 3.5 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) 1.4 m/s and (c) 2.1 m/s?

Explanation / Answer

(a)

The rope does work on the gravity when the skier is pulled up along the incline. as it does work, the kientic energy of the skier change.

Thus, the work done by the rope on the skier is equal to the work done on the skier as the skier moves a distance of 3.5 m up the incline.

It is given that the work done  as the skier moves a distance of 3.5 m up the incline is 750 J.

Therefore, the work done on the skier as moved a distance of 3.5 m up the incline is 750 J.

(b)

It si given that the speed of the rope is constant. This means that the power is also constant.

Power is

P = Work done / time taken

= W / t

= W / (d/v)

= Wv/d

Here, disatnce moved by the skier is d and speed of the rope is v.   

Substitute 750 J for W, 3.5 m for d and 1.4 m/s for v in the above equation,

P =Wv / d

= (750 J)(1.4 m/s) / 3.5 m

= 300 W

Therefore, the rate at which the force of the rope does work on the skier when the rope moves with a speed of 1.4 m/s is 300 W.

(c)

Power is

P = Wv/d

Here, disatnce moved by the skier is d and speed of the rope is v.   

Substitute 750 J for W, 3.5 m for d and 2.1 m/s for v in the above equation,

P =Wv / d

= (750 J)(2.1 m/s) / 3.5 m

= 450 W

Therefore, the rate at which the force of the rope does work on the skier when the rope moves with a speed of 2.1 m/s is 450 W.

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