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Ch 05: Applying Newton's Laws

Chapter 5, Problem 5

A 45.0-kg crate of tools rests on a horizontal floor. You exert a gradually increasing horizontal push on it, and the crate just begins to move when your force exceeds 313 N. Then you must reduce your push to 208 N to keep it moving at a steady 25.0 cm/s. (a) What are the coefficients of static and kinetic friction between the crate and the floor?

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Welcome back everybody. We have a block that is resting on the bed of the truck and we are told that the mass of this block is kg. Now we there's going to be an increasing horizontal force applied to this thing and we are told that there is friction acting in the opposite direction. Now, with any object as well, you're going to have your normal force and your force due to gravity. Now we are tasked with finding what the coefficient of static friction is and the coefficient of kinetic friction is. So first and foremost, let's just start out with Newton's second law here, write Newton's Second law states that the sum of all forces in a given direction. We're gonna look at the extraction in this case is equal to our mass times our acceleration. Well, the forces in the extraction are are applied force minus our force due to friction. And this is equal to Well, in both the case of static friction where the box is not moving and kinetic friction where there is a constant velocity, the right side is going to equal zero, meaning that our applied force is equal to our force due to friction. What is what is the formula for the force due to friction? That is just your coefficient of friction, which with which with ever sorry, with whichever force due to friction. We are looking at times our normal force. Well, what is our normal force? The only forces acting in the y direction are our normal force and are forced to do gravity. So they must be equal meaning that are forced due to friction. New Times our mass times our gravity dividing both sides of the equation here by our mass times the acceleration due to gravity. We get that a given coefficient of friction is equal to our applied force divided by mass times acceleration due to gravity. Now that we have this formula, let's look at our coefficient of friction in both static and kinetic friction. So our coefficient of friction are static, coefficient of friction is able to our force. Our static force, quote unquote divided by our mass due to gravity. Well, what is our static force? Well, we are told that it's going to stay still unless a force greater than 100 and 86 newtons is applied, meaning that we can go up to 100 86 and the box still not move. So we'll have 186 divided by the mass times the acceleration due to gravity which is 9.8. This is equal to 0.76. Giving us our coefficient of static friction. Now, let's look at our coefficient of kinetic friction. This is equal to our kinetic force, quote unquote divided by our mass times acceleration due to gravity. What's our kinetic force? We were told that we will have a constant velocity allowing for this equation to be true when we are pulling with a force of 114 newtons. Now this divided by 25 which is our mass times acceleration due to gravity 9.8 gives us a coefficient of kinetic friction of 0.47. Now we have found our coefficient of static friction and our coefficient of kinetic friction corresponding to answer choice. Thank you all so much for watching. Hope this video helped. We will see you all in the next one.
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A 45.0-kg crate of tools rests on a horizontal floor. You exert a gradually increasing horizontal push on it, and the crate just begins to move when your force exceeds 313 N. Then you must reduce your push to 208 N to keep it moving at a steady 25.0 cm/s. (b) What push must you exert to give it an acceleration of 1.10 m/s2?
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