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$\bullet$ A box of bananas weighing 40.0 $\mathrm{N}$ rests on a horizontal sur-face. The coefficient of static friction between the box and thesurface is $0.40,$ and the coefficient of kinetic friction is $0.20 .$(a) If no horizontal force is applied to the box and the box is atrest, how large is the friction force exerted on the box? (b) Whatis the magnitude of the friction force if a monkey applies a hor-izontal force of 6.0 $\mathrm{N}$ to the box and the box is initially at rest?(c) What minimum horizontal force must the monkey apply tostart the box in motion? (d) What minimum horizontal forcemust the monkey apply to keep the box moving at constantvelocity once it has been started?

A. If there is no applied force, no friction force is needed to keep the box atrest.(b)$f_{s-\max }=\mu_{s} n=(0.40)(40.0 N)=16.0 N$If a horizontal force of 6.0 $\mathrm{N}$ is applied to the box, then $f_{s}=6.0 \mathrm{N}$ in theopposite direction.C. 16.0 $\mathrm{N}$D. 8.0$N$

Physics 101 Mechanics

Chapter 5

Applications of Newton's Law

Motion Along a Straight Line

Motion in 2d or 3d

Newton's Laws of Motion

Applying Newton's Laws

Cornell University

Rutgers, The State University of New Jersey

University of Washington

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question 31 states Any box of bananas wing 40 Newton's rests on a horizontal surface. The coefficient of static friction between the box and the surface is 0.4, and the coefficient of kinetic friction is point to a asks if no horizontal force is applied to the box and the box is at rest. How large is the friction force exerted on the box? Okay, so we'll get to this one first. This scenario, we'll drive. We simply just have a box on the ground, given its weight. And we know both the coefficients of friction. If no force is applied to this box, I mean, there's nothing to no frictional force that would oppose the motion. That means that in this scenario, since there's no force to the thing will oppose that force. So are our frictional force. In this scenario, call just f f has to equal zero as well. Great Part B. What is the magnitude of the friction force if a monkey applies a horizontal force of six minutes to the box and is initially at rest? So now, in part being, we have a a force, um, being pushed on the box which is now six Newtons. So one thing we do know that if we look at just take a free body diagram example of our box, we have the normal force pushing up on the weight of the box, pushing down and for pushing to the right there positive directions they call upon the next. The upward is positive. Why? It means they're frictional force will be opposing Call that little f f frictional force would posing the direction of pushing. So in this scenario, since our force is less than our weight, that means that because we're doing, we do have to deal with, um, there's a friction way to deal with. Um, this force does not overcome the weight of the box. So therefore, the frictional force was something be the amount of force applied on the box, which is simply just 6.0 Newtons such that there will be no movement of the box. So because the force applied does not overcome the force of friction. Do you do this static coefficient of friction? Then simply Morgan, because you know, force of friction is for static scenarios in us times and equals M G tires 40. I'm sorry. US times of wheat. We'll get our wide direction. Our forces. So in order to overcome being pushed, the amount of force that must be applied has 2.4 times the wheat. Just 16 news, which I think, actually has jumped ahead. Yes, this is party. So, Barbie, I justified that because the force being applied is less than the weight. And because friction is something we need to consider the frictional forces. Just six moons. Where in bird, see what minimal horizontal force must the monkey apply to start the box motion? Like I said, we have to overcome the force of static friction. What we found to be 16 mutants and party. What middle of horizontal force must the monkey applied to keep the box moving at a constant velocity? Well, it means we need deal there. Uh, kinetic frictional force. I'm just giving us and UK times the normal forests which based on her free by a diagram. Just nuke eight times the weight. Thank you. Here's point to there was a way of 40 Newtons. What results in a necessary force to push the box of 8.0 news

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