Power question

1. Nov 20, 2008

brittkub1291

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?

2. Nov 20, 2008

tiny-tim

Hi brittkub1291!

Show us what you've tried, and where you're stuck, and then we'll know how to help.

3. Nov 21, 2008

brittkub1291

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.

4. Nov 21, 2008

tiny-tim

Hi brittkub1291!

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

5. Nov 21, 2008

brittkub1291

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.

6. Nov 22, 2008

tiny-tim

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!!! but shouldn't the c be different from the b?

7. Nov 23, 2008

brittkub1291

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...

8. Nov 24, 2008

tiny-tim

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?

9. Nov 24, 2008

brittkub1291

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. Nov 25, 2008

tiny-tim

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

and therefore does not depend on velocity.

11. Nov 26, 2008