Work-energy theorem and resistive forces

In summary: I'm not sure how to approach this problem.In summary, the problem involves a skier starting from rest at a height of 250m and sliding down a hill with an intermediate hill of 100m. The question asks for the speed of the skier at the top and bottom of the intermediate hill, and for the amount of work done by resistive forces of friction and drag when the skier reaches the bottom of the hill with a speed of 28m/s. The work-energy theorem is used to calculate the work done, by subtracting the actual final kinetic energy from the expected final kinetic energy to determine the mechanical energy lost.
  • #1
RedDanger
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Homework Statement


A skier slides down a hill, starting from rest at a height of 250m above the bottom of the hill. She skis over an intermediate hill, whose height is 100m above the bottom of the hill. If resistive forces are neglected, what is the speed of the skier a) at the top of the intermediate hill, b) at the bottom of the hill? c) Suppose the skier reaches the bottom of the hill with a speed of 28m/s. Assuming that the skier, including equipment, has a mass of 85Kg, how much work was done by the resistive forces of friction and drag?


Homework Equations


KEi + PEi = KEf + PEf


The Attempt at a Solution


Parts A and B I don't have trouble with, as they are simply applications of the work-energy theorem. For part A I got 54m/s, and part B I got 70m/s, but part C I have no idea how to approach.
 
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  • #2
Hi RedDanger! :smile:
RedDanger said:
… c) Suppose the skier reaches the bottom of the hill with a speed of 28m/s. Assuming that the skier, including equipment, has a mass of 85Kg, how much work was done by the resistive forces of friction and drag?

Use the work-energy theorem: work done = loss of mechanical energy.

In other words, subtract the actual final KE from the expected final KE … that gives you the mechanical energy lost. :wink:
 

1. What is the work-energy theorem?

The work-energy theorem states that the net work done on an object is equal to the change in its kinetic energy. In other words, when a force acts on an object, it either increases or decreases the object's speed, and this change in speed is directly related to the work done by the force.

2. How does the work-energy theorem apply to resistive forces?

When resistive forces, such as friction or air resistance, act on an object, they do negative work, meaning they decrease the object's kinetic energy. This means that the net work done on the object is equal to the work done by the resistive forces, resulting in a decrease in the object's speed.

3. Can the work-energy theorem be applied to both linear and rotational motion?

Yes, the work-energy theorem can be applied to both linear and rotational motion. In linear motion, the work done by a force is equal to the force multiplied by the distance the object moves in the direction of the force. In rotational motion, the work done by a torque is equal to the torque multiplied by the angular displacement of the object.

4. How do resistive forces affect the efficiency of a machine?

Resistive forces can decrease the efficiency of a machine by doing negative work and reducing the amount of useful work done by the machine. This is because some of the energy input into the machine is lost to overcoming resistive forces instead of being used to do work.

5. How does the work-energy theorem relate to the conservation of energy?

The work-energy theorem is a specific application of the law of conservation of energy, which states that energy cannot be created or destroyed, only transferred from one form to another. In the case of the work-energy theorem, the work done on an object is equal to the change in its kinetic energy, thus demonstrating the conservation of energy.

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