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(II) $\mathrm{A}$ 75-kg snowboarder has an initial velocity of5.0 $\mathrm{m} / \mathrm{s}$ at the top of a $28^{\circ}$ incline (Fig. $36 ) .$ After slidingdown the 110 -m long incline (on which the coefficient ofkinetic friction is $\mu_{k}=0.18$ ), the snowboarder hasattained a velocity $v .$ The snowboarder then slides alonga flat surface (on which $\mu_{k}=0.15$ and comes to restafter a distance $x .$ Use Newton's second law to find thesnowboarder's acceleration while on the incline andwhile on the flat surface. Then use these accelerations todetermine $x .$

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$a_{\text {slope}}=3.04 \mathrm{m} / \mathrm{s}^{2} \quad a_{\text {flat}}=-1.47 \mathrm{m} / \mathrm{s}^{2} \quad x_{f}=235 \mathrm{m}$

Physics 101 Mechanics

Chapter 5

Using Newton's Laws: Friction, Circular Motion, Drag Forces

Motion Along a Straight Line

Motion in 2d or 3d

Newton's Laws of Motion

Applying Newton's Laws

Rotation of Rigid Bodies

Dynamics of Rotational Motion

Equilibrium and Elasticity

Cornell University

University of Washington

Simon Fraser University

University of Winnipeg

Lectures

02:34

In physics, a rigid body is an object that is not deformed by the stress of external forces. The term "rigid body" is used in the context of classical mechanics, where it refers to a body that has no degrees of freedom and is completely described by its position and the forces applied to it. A rigid body is a special case of a solid body, and is one type of spatial body. The term "rigid body" is also used in the context of continuum mechanics, where it refers to a solid body that is deformed by external forces, but does not change in volume. In continuum mechanics, a rigid body is a continuous body that has no internal degrees of freedom. The term "rigid body" is also used in the context of quantum mechanics, where it refers to a body that cannot be squeezed into a smaller volume without changing its shape.

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In physics, rotational dynamics is the study of the kinematics and kinetics of rotational motion, the motion of rigid bodies, and the about axes of the body. It can be divided into the study of torque and the study of angular velocity.

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slipped. Other people diagram off the snow water when he's on the slope. So that's the snowboarder. We see that we have MG that's acting downwards. And then there is the normal force, which is acting particularly which is acting perpendicular to the slow. Let's call it F n. Then there's friction acting along the slope in the upward direction we call it F off fr on DA. For our convenience, let's actually assumed that the perpendicular direction to the slope as positive way and down the slope as positive x so on da If the slope makes angle theater with the or is gentle using trick, we can say that that that angle Tatum was weak will do, uh, that that angle must be equal toe the angle between MG and y axis. So, using Newton's second law, we can write down to several questions one in the extraction, one in the white direction. So the wind direction. We have FN minus mg. Well, Santa is equal to zero, which gives us f N is equal to M G science data and ah, in the words of skidding ray, see, that force of friction is equal to study coefficient of static friction Times F off n where f off in is already calculated here so we can use these two piece of pieces of information in our second part of the question will be sold for ethics. So if X is equal to M G course science data minus f if our Zico to m A where m a so M d co sign today is basically the component of mg along X axis and effort is directed in the opposite direction. That's where you have a negative sign here. So again, as we said that we can combine if and and if are together in this piece on duh, we see that finally we have mg co scientist minus mu s mg Ah, So that's gonna be signed state, actually, because along extraction it's signed. So that's scientific. Similarly, this would be saying as well So in the Sunday that minus mu s mg course, Santa is it will do sometimes a And from here this is the ex elation in the slope. So a slope is equal to D off sign Fade on minus mu ke one f open. Um so that becomes 9.8 meters per second squared times scientist. I would just signed 28 degrees, minus 0.218 course I in 28 degrees, which is equally true. 3.0 40 meters per second squared. So here's I did one mistake here while ah calculating the friction Force it. It's not the static friction, but it's the kind of friction that were considering because that's nobody is moving. So when when someone when an object this movie, we consider the kind of friction kind of the prohibition of panic, friction on deck, a vision of static friction. So that's where we have Mu K one on da Mu K one is denoting that That's the kind of prohibition of kind of friction in the slope. Now we can do the same thing for the second part where the snowboarder is on a flat surface. So if we do so, we see that if the snowblower moves on the right, the friction friction forces Ah, acting on the left. So we do the same thing in the vibration we have ah fn which is counterbalanced by M g. So that's what Stephanie is equal to m. G. And here the frictional forces equal toe condition of kind of friction to where mutate to means. It's on the flat surface times F open. Now we can use these two pieces of information in second part are in in the direction off X. We're forces equal the negative f off afar, which is equal to M times A. Where is the expiration in the flat surface? So that's why we are writing a as a fleck. So combining these three questions, we see that a flat is equal to negative mieux que tu dime. See, we're putting all those numbers. We see that if that becomes negative 1.47 meters per second squared now for the final part, we can use the question 12 point to see 12 twelve two point Well, see to find the speed of the speed at the bottom of the slope. And, ah, this is a speed at the start of the flat section so we can use ah, two point. We'll see again to find the distance X. So first of all, at the end of the slope, we'll be using a two point we'll see, which is the final velocity minus initial velocity. So final velocity squared minus initial velocity squared is equal to two times X elation times X minus X not with the displacement. So displacement is basically the distance of the slope which is 110 meters. So if we solve for the end of the slope So this is the end of the silk. We have squared off the zero squared, which is the initial velocity squared plus two times X elation times X minus X Not so combining everything we see there at the end of the slope. The velocity is 26 135 meters per second. Now this velocity is the start or the initial velocity at the flat surface. So we'll be using the same equation. But now the flat surface. This is my initial velocity and the final velocity zero. So that's why the becomes zero. But V zero is ah, what we got in the previous calculation with 26 135 meters per second and also ah, if that is given as, um negative one point 47 which we calculated in the first part off our calculations. So from here we see that a flat is by negative 1.472 Using that information, we see that the displacement is Ah, 2 36 meters that in the snow Vogler stops after 2 30 See, traveling to 36 meters in the plant service. Thank you.

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