Skier Work and Power Calculation

In summary, a skier is pulled by a tow rope up a frictionless ski slope at an angle of 12° with the horizontal. The rope moves at a constant speed of 1.0 m/s and does 940 J of work on the skier over a distance of 8.8 m. When the rope moves at a speed of 2.0 m/s, the work done by the force of the rope on the skier remains the same. The force of gravity on the skier is mgd sin12º and does not depend on velocity.
  • #1
brittkub1291
24
0
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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  • #2
Hi brittkub1291!

Show us what you've tried, and where you're stuck, and then we'll know how to help. :smile:
 
  • #3
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
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:
 
  • #5
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
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?
 
  • #7
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
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:
 
  • #9
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 :]
 

Related to Skier Work and Power Calculation

1. What is "Skier Power" in terms of physics?

"Skier Power" is a term used to describe the amount of energy or force that a skier is able to generate while skiing. It is influenced by factors such as the skier's mass, speed, and technique.

2. How does the force of gravity affect a skier's power?

The force of gravity plays a crucial role in a skier's power. It is responsible for pulling the skier down the slope and accelerating them. However, the skier can also use their own muscles to generate force and counteract the force of gravity, resulting in more power.

3. How does friction impact a skier's power?

Friction is the resistance between two surfaces, such as the skis and the snow. In skiing, friction is necessary to control and slow down the skier's movements. However, too much friction can also decrease the skier's power by slowing them down and making it harder to move across the snow.

4. What is the relationship between speed and power in skiing?

The skier's speed directly affects their power. A faster speed means a higher kinetic energy, which translates to more power. However, there is a limit to how much power a skier can generate at high speeds, as it becomes harder to maintain control and technique.

5. How do the laws of motion apply to skiing and "Skier Power"?

The laws of motion, specifically Newton's laws, play a significant role in skiing and "Skier Power". The first law states that an object will remain at rest or in motion at a constant speed unless acted upon by a force. In skiing, the skier's power and technique provide the necessary force to overcome the forces of gravity and friction and maintain their motion down the slope. The second law relates force, mass, and acceleration, while the third law states that for every action, there is an equal and opposite reaction. These laws can be observed in a skier's movements and the forces at play while skiing.

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