A box of bananas weighing 40.0 N rests on a horizontal surface. T...

Question

A box of bananas weighing 40.0 N rests on a horizontal surface. The coefficient of static friction between the box and the surface is 0.40, and the coefficient of kinetic friction is 0.20. (e) If the monkey applies a horizontal force of 18.0 N, what is the magnitude of the friction force and what is the box's acceleration?

Accepted Answer

Hey everyone in this problem? We have a 35 newton block resting on a horizontal floor. Okay. The coefficients of static and kinetic friction between the block and floor are 0.38 and 0.26 respectively, a child is pushing against the block with a horizontal force. Okay. We're asked to determine the magnitude of the friction force on the crate and its acceleration when the child a applies a horizontal steady force of 15 newtons. Alright, so let's start by just drawing this out. Okay, so we have our horizontal surface with a block. Okay? We have our child who is pushing the block. Okay, to the right. And if we convert this to a free body diagram, we have our normal force can always pointing up perpendicular to the surface. Okay, I'm opposing that. Or pointing down? We have the weight. W let's take our coordinate system to be the typical coordinate system. Where up into the right is our positive directions. Okay, now this child is pushing the block to the right. Okay, so the force they're applying is going to be to the right. We'll call it F. And were given some information about friction. K. There is friction between the block and the floor, and so the friction is going to oppose the motion. So it's going to act to the left? We have the friction force. Little F. Alright, so the first thing we want to figure out, Okay, when we're looking for the magnitude of the friction force is which friction are we using? Okay. Are we talking about static friction? Or are we talking about kinetic friction? Ok. If the block is still stationary, we're gonna be using static friction. If the block is moving, we're gonna be using kinetic friction. Okay, So let's figure out if this block is moving. Okay? So what we want to do is we want to figure out if F. S. Max the maximum station or static friction force exceeds the force that's being applied, okay or not? So F. S. Max. Well, recall that. This is equal to mu s the coefficient of static friction times N. The normal well, what is the normal? Okay, let's work that out over here. Now, we see the normal force is in the Y direction. So let's consider the sum of the forces in the Y direction. According to Newton's. Second law is going to be the mass times the acceleration in the y direction. The block is not moving up or down, it's resting on a horizontal floor. So the acceleration in the Y direction is zero. Which makes us some of the forces zero. Okay, what are the forces in the Y direction? Well, we have the normal pointing up in the positive Y direction and we have the weight pointing down in the negative Y direction. So we have N minus W. Is equal to zero. We get that the normal force, N is equal to the weight W. And the weight W were given in the problem is 35 newtons. Okay, and just be cautious here, this is an area that you can sometimes mix up. Okay, we're given the block is 35 newtons. Okay, Because we have the unit of newton, that means they're giving us the weight of the block, not the mass of the block. Okay, So it's not 35 kg. It's 35 newtons. Alright, so now we know our normal force. We can go ahead and find this maximum force of static friction. The coefficient of static friction we're given is 0. Times the normal, which is 35 newtons. And we get a maximum sack friction force of 13.3 Newtons. Okay, Alright, so let's compare this with the force that the child is applying. The force the child is applying. F. is 15 Newtons, which is bigger Then F. S. Max Okay, that we just found which was 13.3 Nunes. So what this means is that the child is pushing with enough force to overcome that static friction and therefore the block will be moving. Okay, so we are past the threshold. And so when we're talking about our force of friction, F we're going to be considering F. K. Kinetic friction because that block is moving. Okay, Alright, now that we know what friction force we're talking about, we can calculate, okay, we're trying to calculate the magnitude of the friction force. In this case it's kinetic friction. So let's go ahead and calculate that. Okay, so let's give ourselves some more room. And we're going to say that the force of friction F. Is equal to the force of kinetic friction because that block is moving which is equal to Okay, recall the coefficient of kinetic friction, mu K times the normal N. The coefficient of kinetic friction we're told in the problem is 0.26. The normal we found above to be 35 newtons. And we get this magnitude of the friction force of 9.1 newtons. Okay, Alright, so we've done the first part of the problem. Okay, now the second part of the problem is asking us to calculate the acceleration when the child applies a horizontal study force. Okay, so when we're talking about acceleration, we know through Newton's second law that we can relate that to the sum of the forces. Okay, in this block again we know it's not accelerating in the Y direction. It's going to be accelerating in the X direction because it's resting on that horizontal floor. So we're gonna consider the X. Components. Okay, so the sum of the forces in the X direction, it's going to equal the mass times the acceleration according to Newton's law. What are the some of the forces in the extraction? Let's go back to our free body diagram. Okay, In the positive X direction we have the force that the child is applied by pushing F. And in the negative direction we have that friction force. Little F. So our summer forces is going to be big F. The force of the child pushing the block minus the force of friction. And we'll call it F. K. Because we know that it's kinetic friction because the block is moving is equal to the mass times the acceleration. All right. So, we know big F. We know F. K. We just found it. We know or the acceleration is what we're trying to find now. What about the mass? Okay. Were given the weight? Okay, but we're not giving the mass. So let's recall the weight. W is equal to the mass times of gravitational acceleration. G. Okay. Which tells us that the mass M is going to be equal to the weight divided by G. And in this case our weight was 35 newtons In the gravitational acceleration 9.8 m/s squared. Okay, Alright. So we can get back to our equation here. We know that the applied forces. 15 Newtons. The force of kinetic friction is 9.1 newtons. Okay, the mass is going to be 35 newtons divided by 9.8 m per second squared times the acceleration A. X. Which we're looking for. This gives us A X. Is equal to 15 Newtons -9.1 Newtons times 9.8 m per second squared divided by 35 newtons. The unit of newton will divide out, we're left with meters per second squared, which is what we want for an acceleration to our units check out, we get 1.652 m per second squared for our acceleration. Alright? So if we go back up to the top, we're looking at our answer choices. Okay. The magnitude of the friction force? Well, we've determined that this was the kinetic friction and we found that the magnitude of that force is 9.1 newtons. Okay. And the acceleration when the child is pushing that block and applying that steady force is going to be 1.652 m per second squared. And so we have answer D. Okay, the force of friction 9.1 newtons. And the acceleration is approximately 1.7 m per second squared. Thanks everyone for watching. I hope this video helped see you in the next one.

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