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(II) Piles of snow on slippery roofs can become dangerousprojectiles as they melt. Consider a chunk of snow at theridge of a roof with a slope of $34^{\circ} .(a)$ What is the minimumvalue of the coefficient of static friction that will keep thesnow from sliding down? (b) As the snow begins to melt thecoefficient of static friction decreases and the snow finallyslips. Assuming that the distance from the chunk to the edgeof the roof is 6.0 $\mathrm{m}$ and the coefficient of kinetic friction is$0.20,$ calculate the speed of the snow chunk when it slides offthe roof. (c) If the edge of the roof is 10.0 $\mathrm{m}$ above ground,estimate the speed of the snow when it hits the ground.

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a. $u=0.67$b. $=6.8 \mathrm{~m} / \mathrm{s}$c. $t = 1.42 \mathrm{sec}$

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

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(II) Piles of snow on slip…

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Two Snowy Peaks Two snowy …

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A brick of mass $1.0 \math…

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in the first part of this problem, you're given angle 34 degrees and it says that it is the something for the coefficient of static friction that will keep the snow from sliding. So my maximum static friction is gonna be proportional to my component of gravity That is down the room. So my maximum will be when my MG signed data is equal to my static friction. Now, my static friction maximum is my coefficient of static friction times my normal fours, which in this case, my normal force is equal to the perpendicular component of gravity, which is mg. Oh, Cynthia. Here my mass is canceled. My forces grab my acceleration due to gravity councils and I'm left with the trig function Tangent data equals my coefficient of friction. It's a tangent of 34 gives us eight coefficient of friction of 80.6 seven party for part B. It says that it starts to melt So we want to know the speed in order to know the speed we need first need to know my acceleration. So my acceleration can come from the sum of the forces in the ex direction down the ramp. So we have mg science data minus my force of friction, which is my coefficient temperament, normal force. But now it's the coefficient of kinetic friction, not static friction. So this is mu k. Times mg co signed data equals and A and we can use this to solve for my acceleration. So I have g signed data minus mu k g co signed data. So we have 9.8 times the sign off 34 four minus my new coefficient of friction would just 0.2 9.8 consign off 34 when we get an acceleration of six point by 3.85 and then you can use a cinematic equation to solve for the speed so you need to be squared equals being not squared, plus two A built ex assuming it started from rest. And my final velocity is two times my acceleration times the length that it traveled, which was six meters in the problem, and I get a final velocity of 6.8 meters per second and then knowing my velocity 6.8, I can use that to solve for my velocity as it comes off of the rent. So it's going to go off of our table horizontally and then it's going to be projectile motion after that. So using kid a matic equation I can solve or the time it takes to hit the ground. Initial velocity zero in the UAE direction. So two adults Why, over any acceleration due to gravity gives me a total time in the air of ah, one point for two seconds. If I know the time is one point for two seconds, I can solve for my final velocity and the wind direction, which is 14 meters per second, and then because these are vectors and I have a velocity of 14 meters per second going down and I had a velocity of 6.8 going horizontally than the factory and serum of these two values. Give me my speed in which it hits the ground, which is a 15.56

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