Question

A skier is pulled by a towrope up a frictionless ski slope that makes an angle of 12 degrees with the horizontal. The rope moves parallel to the slope with a constant speed of .0m/s. The force of the rope does 900J of work on the skier as the skier moves a distance of 8.0m up the incline.(a) If the rope moed with a constant speed of 2.0m/s, how much work would the force of the rope do on the skier as the skier moved a distance of 8.0m p the incline? At what rate is the force of the rope doing work on the skier when the rope mves with a speed of
(b) 1.0m/s and
(c) 2.0m/s?

Answer 1

Answer:

Explanation:

Given,

• Work done by the rope 900 m/s.
• Angle of inclination of the slope = $\theta\ =\ 12^o$
• Initial speed of the skier = v = 1.0 m/s
• Length of the inclined surface = d = 8.0 m

part (a)

The rope is doing the work against the gravity on the skier to uplift up to the inclined surface. Therefore the work done by the rope is equal to the work done on the skier due to the gravity

$\therefore W_r\ =\ W_g\ =\ 900\ J$

In both cases the height attained by the skier is equal. and the work done by gravity does not depend upon the speed of the skier.

part (b)

• Initial speed of the skier = v = 1.0 m/s.

Rate of the work done by the rope is power of the rope.

$Power\ =\ \dfrac{\Delta W}{\Delta t}\\\Rightarrow P\ =\ \dfrac{\Delta W}{\dfrac{d}{v}}\\\Rightarrow P\ =\ \dfrac{\Delta W\times v}{d}\\\Rightarrow P\ =\ \dfrac{900\times 1.0}{8.0}\\\Rightarrow P\ =\ 112.5\ Watt$

Part (c)

• Initial speed of the skier = v = 2.0 m/s.

Rate of the work done by the rope is power of the rope.

$Power\ =\ \dfrac{\Delta W}{\Delta t}\\\Rightarrow P\ =\ \dfrac{\Delta W}{\dfrac{d}{v}}\\\Rightarrow P\ =\ \dfrac{\Delta W\times v}{d}\\\Rightarrow P\ =\ \dfrac{900\times 2.0}{8.0}\\\Rightarrow P\ =\ 225\ Watt$