Skier Work and Power Calculation

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Homework Help Overview

The discussion revolves around a physics problem involving a skier being pulled up a frictionless slope by a tow rope. The problem includes calculations of work done by the rope and the power exerted at different speeds of the rope.

Discussion Character

  • Mixed

Approaches and Questions Raised

  • Participants explore the relationship between work, force, and distance, questioning how changes in speed affect the work done. Some participants attempt to apply formulas related to power and work but express uncertainty about their application.

Discussion Status

There is an ongoing exploration of the problem, with participants providing insights and corrections to each other's reasoning. Some have suggested that the work done does not depend on the speed of the rope, while others are still trying to clarify their understanding of the relationships between the variables involved.

Contextual Notes

Participants note confusion regarding the problem's wording, particularly the labeling of parts (e.g., two parts labeled 'b'). There is also mention of the need to consider gravitational force and potential energy in the context of the work-energy theorem.

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A skier pulled by a tow rope 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 940 J of work on the skier as the skier moves a distance of 8.8 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 8.8 m up the incline?

(b) At what rate is the force of the rope doing work on the skier when the rope moves with a speed of 1.0 m/s?

(b) At what rate is the force of the rope doing work on the skier when the rope moves with a speed of 1.0 m/s?
 
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Hi brittkub1291!

Show us what you've tried, and where you're stuck, and then we'll know how to help. :smile:
 
Okay so i think i start with P=Fvcos(x), so for the first one i plugged in the force, velocity and cosine of my angle and got the power. This didn't get me the answer, which is probably because I'm trying to find work. So i know that P=dw/dt, but i don't know the time so I'm kind of stuck.
 
brittkub1291 said:
A skier pulled by a tow rope 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 940 J of work on the skier as the skier moves a distance of 8.8 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 8.8 m up the incline?

(b) At what rate is the force of the rope doing work on the skier when the rope moves with a speed of 1.0 m/s?

(b) At what rate is the force of the rope doing work on the skier when the rope moves with a speed of 1.0 m/s?
brittkub1291 said:
… So i know that P=dw/dt, but i don't know the time so I'm kind of stuck.

Hi brittkub1291! :smile:

(I don't know why you need to use power to answer a b and b … uhh, why are there two b's? :confused: … but anyway:-)

Yes, you do know the time … it's the time to go 8.8 m at 1.0 (or 2.0) m/s :redface:
 
Okay so for the first one i tried W=940Ncos(12)*8.8m, but it didnt work... And the second b should be a c lol sorry.
 
brittkub1291 said:
Okay so for the first one i tried W=940Ncos(12)*8.8m, but it didnt work... And the second b should be a c lol sorry.

No, you're not reading the question properly …

940 isn't the force, it's the work done at 1.0 m/s over 8.8m …

(and the force is parallel to the slope anyway)

yes, i worked out it should be a c! :rolleyes: but shouldn't the c be different from the b?
 
Ah, I'm not so good with the cutting and pasting apparently. Here's what it should be:

c. At what rate is the force of the rope doing work on the skier when the rope moves with a speed of 2.0 m/s?

And your right I'm not reading it correctly. So i figure now that since the work is 940 over a distance of 8.8m then the force should be 106.818 using W=F*x. But then i need to figure out what the work is for 8.8m at 2m/s so I'm not sure what to do...
 
brittkub1291 said:
Ah, I'm not so good with the cutting and pasting apparently. Here's what it should be:

c. At what rate is the force of the rope doing work on the skier when the rope moves with a speed of 2.0 m/s?

And your right I'm not reading it correctly. So i figure now that since the work is 940 over a distance of 8.8m then the force should be 106.818 using W=F*x. But then i need to figure out what the work is for 8.8m at 2m/s so I'm not sure what to do...

Let's do this logically …

you have W d and v (for work distance and speed) …

(note: you're not given the force or the time, and the question doesn't ask for them)

i] write out a formula using only W d and v

ii] looking at the formula, what happens to W if we change v but keep d the same? :wink:
 
Okay i think i figured it out. The force of gravity on the skier should be mgd and then the force of the rope on the skier should be the same. So the velocity doesn't matter, the work is the same either speed.
 
  • #10
brittkub1291 said:
Okay i think i figured it out. The force of gravity on the skier should be mgd and then the force of the rope on the skier should be the same. So the velocity doesn't matter, the work is the same either speed.

hmm … more-or-less right, but not a logical way of saying it.

Work done = change in energy (work-energy theorem)

= change in PE (because KE is constant)

= mgd sin12º :wink:

and therefore does not depend on velocity.
 
  • #11
Thanks for your help :]
 

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