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Multiple-Concept Example 5 reviews many of the concepts that play roles in this problem. An extreme skier, starting from rest, coasts down a mountain slope that makes an angle of $25.0^{\circ}$ with the horizontal. The coefficient of kinetic friction between her skis and the snow is 0.200. She coasts down a distance of 10.4 m before coming to the edge of a cliff. Without slowing down, she skis off the cliff and lands downhill at a point whose vertical distance is 3.50 m below the edge. How fast is she going just before she lands?

$v=10.9 \mathrm{~m} / \mathrm{s}$

Physics 101 Mechanics

Chapter 6

Work and Energy

Work

Kinetic Energy

University of Michigan - Ann Arbor

Simon Fraser University

Hope College

University of Winnipeg

Lectures

03:47

In physics, the kinetic energy of an object is the energy which it possesses due to its motion. It is defined as the work needed to accelerate a body of a given mass from rest to its stated velocity. Having gained this energy during its acceleration, the body maintains this kinetic energy unless its speed changes. The same amount of work is done by the body in decelerating from its current speed to a state of rest. The kinetic energy of a rotating object is the sum of the kinetic energies of the object's parts.

02:08

In physics, work is the transfer of energy by a force acting through a distance. The "work" of a force F on an object that it pushes is defined as the product of the force and the distance through which it moves the object. For example, if a force of 10 newtons (N) acts through a distance of 2 meters (m), then doing 10 joules (J) of work on that object requires exerting a force of 10 N for 2 m. Work is a scalar quantity, meaning that it can be described by a single number-for example, if a force of 3 newtons acts through a distance of 2 meters, then the work done is 6 joules. Work is due to a force acting on a point that is stationary-that is, a point where the force is applied does not move. By Newton's third law, the force of the reaction is equal and opposite to the force of the action, so the point where the force is applied does work on the person applying the force. In the example above, the force of the person pushing the block is 3 N. The force of the block on the person is also 3 N. The difference between the two forces is the work done on the block by the person, which can be calculated as the force of the block times the distance through which it moves, or 3 N × 2 m = 6 J.

05:08

Multiple-Concept Example 5…

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10:02

An extreme skier, starting…

04:37

$\bullet$A skier approache…

05:26

A skier leaves the startin…

14:46

Use the work$-$energy theo…

09:02

A skier leaves the ramp of…

03:26

A child slides down a snow…

02:40

A skier starts at rest ato…

04:52

A 60.0 -kg skier sturts fr…

08:00

A 60.0-kg skier starts fro…

So let's start by making a picture here, so we have a inclined plane. The angle off in commission is 25 degrees at the end. Off the incline, there's there's a cliff and then s Oh, there's horizontal drunk. So our skier starts right here, and at this point the initial velocity is zero. And then along this inclined plane, there's going to be friction. So are our skier is going to slide down and then at the edge off at drop off at the edge here. And then once these gear jumps off, the path would be a problem. A problem. The length off this incline is 10.4 meters. So let's first of all figured out. What is the vertical height here and that vertical height. Let me call that each one would be cool to 10.4, sign off 25 degrees and that comes out to be equal to 4.4 meters. And the second height here is 3.5 meters, which means that the overall height through which the skier will fall is going to be 7.9 meters. That's going to be important. Now we know that the Delta K or the change in kinetic energy is equal to the network. Done So we have 1/2 em V squared minus V, not squared is equal to now the work done by gravity. That's going to be one thing. And then the work done by friction. That's going to be another thing. Plus the work done by the normal force. Now the work done by normal force is zero because the normal force would be perpendicular to the motion. So initial velocity is also zero. So I can actually simplify this. To say 1/2 m b squared is equal to work under gravity, which would be MGH, and he worked on by friction would be a negative work because thief force of friction is opposite to the direction off motion. So I can say that's going to F. K s. So now we have 1/2 and B squared is equal to em. G times 7.9 minus mu K Times MG CO sign off theta times es and S s 10.4. So one thing we can do to simplify things, it's actually get rid of the M in each off those terms so we have 1/2 V squared is equal to 9.8 times 7.9 minus 0.2 times 9.8 times course, sign off 25 degrees and then times 10.4. So when I saw this for V, the value of me I get is 10.9 meters per second.

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