Work of Rope on Skier on 12° Incline: 900 J or 2,000 J?

In summary: I'm sorry if I'm not making myself clear. The Work-Energy theorem is W_{total} = W_{net} = \Delta KE, where W_c is the work done by conservative forces (like gravity), and W_{nc} is the work done by non-conservative forces (like friction or applied forces). The Conservation of Total Energy principle can be written W_{nc} = \Delta KE + \Delta PE. You can use either formula, just be careful how you define your variables. The latter is a bit easier to work with in this case.
  • #1
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Homework Statement


A skier is pulled up 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 1.0 m/s. The force of the rope does 900 J of work on the skier as the skier moves a distance of 8m up the incline. If the rope moved with a constant speed of 2 m/s, how much work would the force of the rope do on the skier as the skier moved a distance of 8m up the incline?

Homework Equations


W=KE or W=KE+PE


The Attempt at a Solution



This is a rather conceptual problem in my opinion. So I know the work-kinetic energy theorem. change in K=W. However, my question is: Does it only apply to a system where there is only kinetic energy present or it can be applied to a system where both Kinetic energy and potential energy present. I didn't use K=W, I had W=KE+PE. KE is 0, so W=PE. Because in this case, clearly, there are both Kinetic energy and Potential energy (8m) on the skier. But my teacher still used change in K=W=0, and since the work includes the work from the rope and gravity, they must be same in magnitude and opposite in direction. I hope someone can clarify things for me. thanks.
 
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  • #2
If you're carrying something up a hill, working against a force like gravity, your work is equal to the potential energy you have to overcome to move against the field. If you pushed something up a hill so hard that you increased the velocity of it, you'd have a kinetic term to add, because changing the velocity of something requires an acceleration (note that when you referenced your teacher you spoke of a change in velocity).

Remember Newton's First Law. It implies that keeping and object at a constant velocity does not require extra work theoretically (in the real world, we're constantly battling friction, so we have to put extra energy into things to keep them going).
 
  • #3
Please be careful with your subscripts when takling about "Work". The Work-Energy theorem is [tex]W_{total} = W_{net} = \Delta KE [/tex], where
[tex]W_{total} = W_c + W_{nc}[/tex], and where [tex]W_c[/tex] is the work done by conservative forces (like gravity), and [tex]W_{nc}[/tex] is the work done by non-conservative forces (like friction or applied forces).
Now the general conservation of total energy principle can be written [tex] W_{nc} = \Delta KE + \Delta PE[/tex]. You can use either formula, just be careful how you define your variables. The latter is a bit easier to work with in this case.
 
  • #4
Hi, Jay. So change in KE is the same as NET WORK. and the conservation of energy equation: W=change in KE+ change in PE that we often use is for W for non-conservative force like applied forces? Never realized this before. When I learned energy, I first learned change in KE equals to work, but then I learned W=KE+PE. So if there is PE in a problem, we can't just write W=change in KE, we have to include PE. According to you, if we don't have gravity, then W total=Wnc=change in KE? But if there is no gravitation, then there is no PE either?
 
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Related to Work of Rope on Skier on 12° Incline: 900 J or 2,000 J?

1. How is the work of rope on a skier on a 12° incline calculated?

The work of rope on a skier on a 12° incline is calculated using the formula W = Fdcosθ, where W is the work done, F is the force applied by the rope, d is the distance traveled by the skier, and θ is the angle of the incline.

2. How does the angle of incline affect the work of rope on a skier?

The angle of incline affects the work of rope on a skier because it determines the component of the force applied by the rope that acts in the direction of motion. The steeper the incline, the more the force is directed against the direction of motion, resulting in a lower work done by the rope.

3. Is the work done by the rope on a skier on a 12° incline the same as the work done by the skier?

No, the work done by the rope on a skier on a 12° incline is not the same as the work done by the skier. The rope only exerts a force on the skier, while the skier also exerts a force on the rope. This results in a net work done by both the rope and the skier.

4. Can the work of rope on a skier on a 12° incline be negative?

Yes, the work of rope on a skier on a 12° incline can be negative. This means that the force applied by the rope is acting against the direction of motion, resulting in a decrease in the skier's energy. This can happen when the skier is moving downhill and the rope is pulling them uphill.

5. How does the work of rope on a skier on a 12° incline affect the skier's kinetic energy?

The work of rope on a skier on a 12° incline can either increase or decrease the skier's kinetic energy, depending on the direction of the force applied by the rope. If the force is in the same direction as the skier's motion, it will increase their kinetic energy. If the force is in the opposite direction, it will decrease their kinetic energy.

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