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. (d) What minimum horizontal force must the monkey apply to keep the box moving at constant velocity once it has been started?

Accepted Answer

Hey everyone in this problem, we have a 42 Newton block resting on a horizontal floor. The coefficient of static and kinetic friction between the block and the floor are 0.46 and 0.29, respectively. A child is leaning against the block applying a horizontal force and were asked to determine the least horizontal force the child must apply to move the crate at a constant velocity. Okay, Alright, so we draw a little diagram here, we have the floor, we have the block here. Okay, We have a child leaning up against this block, causing it to move with some constant velocity. V. Alright, so if we convert this into a free body diagram for a walk, we know that we have the normal force pointing up, Okay, perpendicular to the surface, we have the weight w pointing down. Okay, we're gonna take up into the writers are positive directions, X and Y. K. Now this trial is leaning against the block, it's going to push the block to the right, and so we have our applied force F pointing to the right, and if you draw drew this the opposite direction where you had the child pushing the block to the left. That's okay too. Um It's just going to change the signs around, but the answer will work out the same in the end. Now, we're also told we have friction. Okay, so if our applied forces to the right, our block is moving to the right friction is going to oppose or resist. That force, or friction force is going to be pointing to the left friction force. F. Okay, and this is gonna be F. Sub K. It's gonna be the force of kinetic friction. We know we have kinetic friction because this box or the crate is moving. Okay, we're told it has a constant velocity it's moving. So we're gonna have kinetic friction. Alright, so we want to determine that least horizontal force. Okay, So we want to determine F. We know that F. Is in our X. Direction. So let's start with our X components. Okay. Newton's second law tells us that the sum of the forces in the X direction is equal to the mass times the acceleration in the extraction and again we have constant velocity. Okay, when we have constant velocity that means the acceleration is zero. And so the sum of the forces is going to be zero. Okay, what is our sum of forces? Okay, well in the extraction, the positive extraction we have our applied force F. And in the negative extraction we have our force of friction kinetic friction F. K. So we have big F minus little F. Sub K. Is equal to zero. Well this tells us that are applied force F must be equal to our kinetic friction force F. K. And recall that the force of kinetic friction is given by mu K. The coefficient of kinetic friction times end the normal, we know mu K. Were given in the problem but we don't yet know the normal force. Okay, so let's go over to the y direction where we have this normal force and see if we can determine what N. Is. Okay, so similar to the X direction, in the Y direction. Newton's second law tells us we have the sum the forces in the Y direction equals M. A. Y. Mass times acceleration in the Y direction. The block or the crate is not moving up and down, in case the acceleration in the Y direction is zero, sum of the forces is zero. Okay, our forces in the Y direction we have in the positive Y direction, the normal force end and in the negative Y direction, the weight. W. So we have N minus W equals zero. So the normal is equal to the weight. And we're told that this block is 42 newtons. So we're told to wait. Okay, so the normal end is just going to be equal to 42 newtons and just watch there. That's a common mistake. Okay, In this case we're given 42 newtons for the block, which means that that's the blocks. Wait, Okay, We're not given its mass. Were not given 42 kg were given 42 newtons. So that's its weight, not its mass. Alright, so, we know our normal force and we can get back over here to find our applied force F. And we get that it is mu K coefficient of kinetic friction, which is 0.29. Given in the problem is the normal force 42 newtons, which we just found, which gives us a force of 12.18 newtons. Okay. And so that least horizontal force that the child must apply to have that crate moving at a constant velocity is 12.18 newtons. And if we look at our answer choices, we see that that's going around to 12.2 newtons. Answer be. Thanks everyone for watching. I hope this video helped see you in the next one.

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