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ssm A 6.00-kg box is sliding across the horizontal floor of an elevator. The coefficient of kinetic friction between the box and the floor is 0.360. Determine the kinetic frictional force that acts on the box when the elevator is (a) stationary, (b) accelerating upward with an acceleration whose magnitude is 1.20 $\mathrm{m} / \mathrm{s}^{2}$ , and $(\mathrm{c})$ accelerating downward with an acceleration whose magnitude is 1.20 $\mathrm{m} / \mathrm{s}^{2}$ .

21.2 $\mathrm{N}$ 23.8 $\mathrm{N}$ 18.6 $\mathrm{N}$

Physics 101 Mechanics

Chapter 4

Forces and Newton’s Laws of Motion

Newton's Laws of Motion

Applying Newton's Laws

Rutgers, The State University of New Jersey

Simon Fraser University

Hope College

University of Sheffield

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we begin this question by setting up our reference frame and then applying Newton's second law to the box. So I will choose my reference frame as follows. Everything that is pointing up is positive and everything that is pointing to the right. It's positive. So as a consequence, everything that is pointing to the left will be negative, and everything that is pointing downwards is also negative. Now we proceed to apply Newton's second law to that box. So let me call these vertical axis the Y axis and this horizontal access access access. So now let us apply mutant second low to the vertical axis to the box. So the net force that is acting on the box on the vertical packs is is equals to the mass off the box times its acceleration in the vertical axis. So in that access, there are two forces acting on the box, the normal force that is pointing to the positive direction on the weight for its that points to the negative direction. And this is the goes to the mass Off the box times the acceleration off that box. No, no to the following note. The following week, uncouple it's normal force as the mass off the box times the vertical acceleration off the box, plus the weight off the box. Now, remember that the frequent oh force that acts on the box the absolute value of that frictional force is given by the genetic frictional coefficient times the normal force. Therefore, the original force is equals to the kinetic frictional coefficient times the mass off the box times the vertical acceleration off the books, plus the weight off this box. Putting in some values that the problem give us nutritional force is in general equals to 0.36 times six, which is the mass off the box times the vertical acceleration off the elevator, plus the weight off the box. Now for the weight. Remember that the weight is given by the mass times acceleration of gravity on that celebration off gravity is approximately nine 0.8 meters per second squared. So we have six, which is the mask off the box times 9.8 to compose the weight off that box. And this is our equation for the frictional force. The situation now all we have to do is apply it or on each item using the respective acceleration. So for the first item, the elevator is a stationary. So in the situation, the vertical acceleration off the box is in close to zero because the elevator is the stationary. So the box is are so stationary on the vertical axis, it still moves on the horizontal axis. Therefore, the frictional force is it close to 0.36 times six times 9.8. This is because this term is close to zero. When the acceleration is it cost zero. So we're left with this term time 0.36. And this gives us a frictional force off approximately 21.2 new terms. So this is the answer off the first item? No. On the second item, we have an acceleration that is pointing upwards. So it's the positive acceleration off 1.2 meters per second squared. Then in the situation, the frictional force Zico's to 0.36 times, six times one point true plus six times 9.8. And these gives a frictional force off approximately 23.8 Muto's So this is the answer off the second item and finally on the third item, we have an acceleration off 1.2 meters per second. But this time the acceleration is pointing downwards, sir. Now reference frame. It's a negative acceleration. So now minus 1.2 meters per second squared. And this gives us a frictional force off 0.36 times, minus six times one point true. Plus six times 9.8. So the frictional force in the situation is approximately 18 0.6 neutrons. So this is the answer for the third and final item.

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