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(II) A child slides down a slide with a $34^{\circ}$ incline, and at thebottom her speed is precisely half what it would have beenif the slide had been frictionless. Calculate the coefficient ofkinetic friction between the slide and the child.

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$\mu=0.5058$

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

Simon Fraser University

Hope College

Lectures

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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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(II) A child slides down a…

(III) A child slides down …

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we have a child, it's lies down a 34 degree incline, and we want to know the cowfish of kinetic friction between the child and the slide. If the child has half the speed when there is friction compared thio when there is not frictions at the bottom of the incline. So first off we start with a free body. Diagram was a component of gravity that goes down the incline in a component of gravity that goes into the pipeline. CO sign date. An Sisk MG signed data and we can write our first equation for when there is no friction. So we have mg signed data again. No friction in this first situation peoples in May. So my acceleration is g signed data and I can get an acceleration of, um, point a 5.48 year again. This is the acceleration the child has when there is no friction. So, using a cinematic the equation, I could say the initial velocity zero So my final velocity squared equals to a deep. Then we could square root that to get a final velocity of square to 80 Now, using the same reasoning initial velocity of zero my final velocity for this second acceleration for when it does have friction is going to be equal to this value divided by two. So you have the spirit of two. Aye, We'll call this one. Do you divided by two, equals the square to a to D. Now here, if you'd like you can What? The two inside of the square root. So that means I would have to mold by this inside part by four to bring that to into the square root. So this becomes ate A to D spit right in. And then we can counsel up a square roots. We can cancel out the distances because that will be the same. And we can reduce this to an eight to a one for a two. So that tells us that my second acceleration for when there is friction is going to be 5.48 divided by four, which gives us 1.37 meters per second squared. So going back to my situation in which there is friction, I can write a new equation minus friction again. In this case, I have the coefficient of kinetic friction times mg coastline data which is my normal force. In this case, masses cancel out and I want to solve for the coefficients of kinetic friction. And then I want to get me by itself and using my second acceleration. This is for when we do have friction, I get a coefficient of friction of zero point five 05

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