Question

A 62 - kg person on skies is going down a hill sloped at 37 degrees. There is 200N of friction resisting this motion. 1. Draw and label the following forces and components acting on the crate: a. Weight (Fg) b. Normal Force (FN) c. Force of Friction = 200N d. Component of weight parallel to plane of motion e. Component of weight perpendicular to plane of motion 2. Solve for the weight of the skier. 3. Solve for the normal force 4. What is the net force in the y-direction? 5. What is the net force in the x-direction? 6. Is this skier accelerating? Explain your answer 

          A 62 - kg person on skies is going down a hill sloped at 37 degrees. There is 200N of friction resisting this motion.
1. Draw and label the following forces and components acting on the crate:
a. Weight (Fg)
b. Normal Force (FN)
c. Force of Friction = 200N
d. Component of weight parallel to plane of motion
e. Component of weight perpendicular to plane of motion
2. Solve for the weight of the skier.
3. Solve for the normal force
4. What is the net force in the y-direction?
5. What is the net force in the x-direction?
6. Is this skier accelerating? Explain your answer
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A 62 - kg person on skies is going down a hill sloped at 37 degrees. There is 200N of friction resisting this motion. 1. Draw and label the following forces and components acting on the crate: a. Weight (Fg) b. Normal Force (FN) c. Force of Friction = 200N d. Component of weight parallel to plane of motion e. Component of weight perpendicular to plane of motion 2. Solve for the weight of the skier. 3. Solve for the normal force 4. What is the net force in the y-direction? 5. What is the net force in the x-direction? 6. Is this skier accelerating? Explain your answer

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University Physics with Modern Physics
University Physics with Modern Physics
Hugh D. Young 14th Edition
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Problem #2: A 62 kg person on skis is going down a hill sloped at 37 degrees. There is 200 N of friction resisting this motion. Draw and label the following forces and components acting on the skier: Weight (Fg), Normal Force (Fn), Force of Friction (Ff), Component of weight parallel to plane of motion, Component of weight perpendicular to plane of motion. Solve for the weight of the skier. Solve for the normal force. What is the net force in the y-direction? What is the net force in the x-direction? Is this skier accelerating? Explain your answer.
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Transcript

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00:01 All right, so let's say we have a skier going down an incline like this.
00:05 And if we draw a free body diagram, the weight of the skier is going to go this way.
00:13 The component of the weight along the plane is going to be m .g.
00:17 Sine theta.
00:18 And the component perpendicular to the plane is m .g.
00:22 Coat -sign theta.
00:23 So this is, you know, the weight is just m .g.
00:29 And the frictional force then is going to act this way.
00:34 I think that's all we need.
00:36 So then for part two, so that was part one, part two, we want to solve for the weight of the skier, which is easy.
00:44 It's just 62 kilograms times 9 .8 meters per second squared.
00:55 So 607 .6 newton's question three.
01:01 What is the normal force? so we'll call it it fn.
01:04 This is just going to be the weight times.
01:06 The cosine of the angle...
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