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(I) A force of 35.0 $\mathrm{N}$ is required to start a 6.0 -kg box movingacross a horizontal concrete floor. $(a)$ What is the coefficientof static friction between the box and the floor? (b) If the35.0 -N force continues, the box accelerates at 0.60 $\mathrm{m} / \mathrm{s}^{2}$ .What is the coefficient of kinetic friction?

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a. =0.6b. =0.53

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

Andres C.

February 23, 2021

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body diagram off this, uh, scenario. We see that we have the box around here and then it's on the floor now, the boxes moving horizontally, Let's say the boxes moving from left. Oh, from left to right. And then because of that, we have some forces that's acting on the books. First of all, we have the force that's acting on the box. Let's call that F B. And since the boxes moving from left to right, there will be some friction force that's acting on the box. And the direction would be opposite because fiction acts in the negative direction or in the opposite direction. Then in vertically vertical direction, we have MG exacting downwards. And then we have the normal force is acting airports. Let's say the boxes moving with an expiration eat. So from left to right, declaration of off the book boxes, eh? Now, in first part, we need to figure out the coefficient of static friction. So for static friction, that means, um, it's, uh it should be equal toe the, um no. Normal. So we know that friction force F R. If that's a static friction than this is gonna be new s times f um, let's call this force end, which is this guy over here. So let's actually call this guy f and s O that we're just consistent with the units, right? Um, and by friction force, it means that before the box starts to move, what is the fictional four that's acting on the block? That's the static friction. So this friction right here is just before the box starts to move. So that means in the, um in the horizontal direction. The box hasn't hasn't been started. Hasn't been started moving yet, so the total force in their direction is zero. So if we recall this left with my direction as X, the total force will be zero. And that's equal to, um f d minus and are because f r is f off fr, which is this way and F p's from left to right. So they're balancing each other when there's a static, which when we're calculating the coefficient of static friction and from here we see that F p is equal to, um if off, if our which is from here, we see its new s times f n where f n can be found using the particle motion where MD is acting downwards and since there is no movement in the vertical direction, that means the normal force must counterbalance this force that's going in the ER vertical direction. So that tells us that F n is equal to M. G. So we can use that relation over here. So that gives us us times, N. G. And from here, if we saw form us, that gives us ft over mg. So F B is 35 mutants and ah m is six kg kinds, 9.8 meters per second squared, which is G and that gives us 0.6 as our coefficient of static friction. So that's part Aye, or one. Whatever you wanna call it now for part B, we have the we have to find out the condition of kind of deflection. And this time the box is accelerating. And if we remember the diagram, we had a box moving from left to right, and the ex elation is given as 0.6 meters per second squared. So now again, if we use the given the questions we see, the only thing that'll change here is the horizontal direction because now the box is moving, so the total force will be m times A Because the box is moving with an expiration A so don't force should be mass times say, On the other hand, we know that there is F b which is acting from left, right? And then there is if off f s, which is acting from right to left. So let's go. It our for the kind of decryption. So yeah, I think Yeah, it's if are so, Yeah, we have f r Ridge is going from right to left. So, uh, will have f b minus f off f R, which is equal to m A. And the reason we're subtracting F off F ar from FT is because the whole system is moving from left to right. That means this force is dominant over this force. Now, if we sold for it again, we can use for this guest Now it's the, um, kind of friction. So this is gonna be UK in sort of new S times f off in. And if we remember this correctly, we have md acting downwards, and then we have an end, which is acting upwards. So from here we see that if off p minus mule K times MG is equal to eight. And from there we see that Mu K is equal to if off b minus. M a divided by mg. Now the given values for given values are for FT. It's 35 Newton's M R D mass. Off the block is six kg e o. The exploration is 00.6. Meet over second squared and G is 9.8. Meet over seconds for And from here. If we plug in all the numbers over here, we see that nuke a is point five p. Thank you.

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