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A skier is pulled by a towrope up a frictionless ski slope that makes an angle of 12 with the horizontal. The rope moves parallel to the slope with a constant speed of 1.0 m/s. The force of the rope does 900 J of work on the skier as the skier moves a distance of 8.0 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 8.0 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) 1.0 m/s and (c) 2.0 m/s?

Respuesta :

Answer

given,

angle with horizontal = 12°

constant speed with the slope = 1 m/s

Work on the skier to move 8 m = 900 J

a) Rope moved with constant speed

change in kinetic energy is equal to zero

 work done by the force of the rope = 900 J

rate of force of the rope when skier move with

b) v = 1 m/s   total time t = 8 s

            Power =[tex]\dfrac{work\ done}{time}[/tex]

                       =[tex]\dfrac{900}{8}[/tex]

                       = 112.5 W

c) v = 2 m/s   total time t = 8/2 = 4 s

            Power =[tex]\dfrac{work\ done}{time}[/tex]

                       =[tex]\dfrac{900}{4}[/tex]

                       = 225 W

ajeigbeibraheem

A skier is pulled by a towrope up a frictionless ski slope that makes an angle of 12 with the horizontal. The rope moves parallel to the slope with a constant speed of 1.0 m/s. If the rope moved with a constant speed of 2.0 m/s, then the workforce the rope will do on the skier is 900 J. The rate at which the force of the rope is doing the work on the skier when the rope moves with a speed of 1.0 m/s and 2.0 m/s are 112.5 watts and 225 watts respectively.

From the given information;

  • The frictional force of the ski slope is = 0 since the slope is said to be frictionless.
  • Similarly, the work done by the rope is equal to the work done by gravity

Now, by applying the kinetic energy K.E and work-energy theorem, we have:

[tex]\mathbf{\Delta K = W_{net}}[/tex]

[tex]\mathbf{K_f - K_i = W_r + W_g}[/tex]

where;

  • [tex]\mathbf{k_f}[/tex] = final Kinetic energy
  • [tex]\mathbf{k_i}[/tex] = initial kinetic energy
  • [tex]\mathbf{w_r}[/tex] = work done by the rope
  • [tex]\mathbf{w_g}[/tex] = work done by gravity

So, if the rope moved with a constant speed of 2.0 m/s, Thus, the change in the K.E is zero.

[tex]\mathbf{0 = W_r + W_g}[/tex]

[tex]\mathbf{ W_g= - W_r = 900 \ J}[/tex]

However, the speed of the rope remains constant because the work done on the skier doesn't increase.

∴

The rate at which the work done on the skier refers to the Power used by the skier which is expressed by using the formula:

[tex]\mathbf{Power = \dfrac{W_r}{t}}[/tex]

where;

  • t = time

Recall that speed = distance/time, So by making time the subject of the formula, we have:

  • time = distance/speed

Given that, the distance to move up the incline = 8.0 m

At the speed of (a) 1.0 m/s

  • time  = 8.0 m / 1 .0 m/s
  • time = 8 s

Now,

[tex]\mathbf{Power = \dfrac{900}{8}}[/tex]

Power  = 112.5 watts when the rope moves with a speed of 1.0 m/s

When the rope moves with a speed 2.0 m/s

  • time = 8.0 m/ 2.0 m/s
  • time = 4 s

Now,

[tex]\mathbf{Power = \dfrac{900}{4}}[/tex]

Power = 225 watts when the rope moves with a speed of 2.0 m/s

Therefore, we can conclude that the workforce the rope will do on the skier is 900 J. The rate at which the force of the rope is doing the work on the skier when the rope moves with a speed of 1.0 m/s and 2.0 m/s are 112.5 watts and 225 watts respectively.

Learn more about work done here:

https://brainly.com/question/13662169?referrer=searchResults

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