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$\bullet$A skier approaches the base of an icy hill with a speed of12.5 $\mathrm{m} / \mathrm{s}$ . The hill slopes upward at $24^{\circ}$ above the horizontal.Ignoring all friction forces, find the acceleration of this skier(a) when she is going up the hill, (b) when she has reached herhighest point, and $(\mathrm{c})$ after she has started sliding down thehill. In each case, start with a free-body diagram of the skier.

a) 4 $\mathrm{m} / \mathrm{s}^{2}$b) 4 $\mathrm{m} / \mathrm{s}^{2}$c) 4 $\mathrm{m} / \mathrm{s}^{2}$

Physics 101 Mechanics

Chapter 5

Applications of Newton's Law

Motion Along a Straight Line

Motion in 2d or 3d

Newton's Laws of Motion

Applying Newton's Laws

Simon Fraser University

Hope College

University of Sheffield

University of Winnipeg

Lectures

04:01

2D kinematics is the study of the movement of an object in two dimensions, usually in a Cartesian coordinate system. The study of the movement of an object in only one dimension is called 1D kinematics. The study of the movement of an object in three dimensions is called 3D kinematics.

03:28

Newton's Laws of Motion are three physical laws that, laid the foundation for classical mechanics. They describe the relationship between a body and the forces acting upon it, and its motion in response to those forces. These three laws have been expressed in several ways, over nearly three centuries, and can be summarised as follows: In his 1687 "Philosophiæ Naturalis Principia Mathematica" ("Mathematical Principles of Natural Philosophy"), Isaac Newton set out three laws of motion. The first law defines the force F, the second law defines the mass m, and the third law defines the acceleration a. The first law states that if the net force acting upon a body is zero, its velocity will not change; the second law states that the acceleration of a body is proportional to the net force acting upon it, and the third law states that for every action there is an equal and opposite reaction.

02:36

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question 27 states that a skier approaches the base of an icy hill with a speed of 12.5 meters per second. The hill slopes upward at 24 degrees above the horizontal, ignoring all friction forces. Find the acceleration of the skier A She's going up the hill be which he reaches when she has reached her highest point and see, after she had started sliding Gather Hill in each case, sort of the free body diagram of the skier. So this scenario she's traveling at on this icy hill inclined at this angle of 24 degrees, Um, given here by Alfa. So the free body diagram for part a fight just represent your of the square. She starts traveling with force M A with direction the bacon, but it will be drifting downward. Um, actually, I'm sure it wasn't access. I should angle it with my, uh what attracted is the same. I should angle it with my, um, my slope to me. Why Commodities directly upward? Um, sorry. It's It's directly upward from the surface of the hill. I'm always downward. Have awake about it in my Y direction here. If my angles alphas given at the base of the uh, the slope in this angle here is also Alfa and there's an X component, um normal for spurs as X component of the wheat. Based on again, If this is Alfa, that means our weight in our X direction is given to be. It's this portion here which is the sine of angles That's wait times sign Alfa. Therefore, our weight and our Y direction that could put it must be the wait times coz I over I'm just trying like the whole This is our free body diagram you want to solve for the acceleration. You know that it forces acting on direction. It's just given my Emmy, which has to eat counter act with the weight in the X direction. Acceleration is simply sorry. This is double W X to W x. Again is w sign. Alfa Acceleration is divided by the mass. However, we can represent weight as just mass. Sometimes G over sign Alfa did by my am. The acceleration in this case is just g times sign of 24 which was occur to be That's right, 4.0 meters per second squared part of the problem when she reaches the highest point. So that scenario, you no longer than force acting this direction, the forces in the same direction downward and all the other components are the same. W force W X Let me. Why so this scenario, our forces acting in the same direction of their, um as our weight in the X direction because they're acting the same. Similarly, if I m A, he goes a siren native exiting the same direction. So that must mean by the same equation. It's just G sign Alfa for this time is negative. And so it's, of course, traveling down the slope at 4.0 meters per second squared and similarly for C as she's, um after she starts laying down, it's the same cereals this she still feels that pulled down the hill, and the rates are acceleration again. Down the hill is four point no meters per second squared

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