To pull a 51 kg crate across a rough floor, a worker applies a force of 100 N, directed 37°above the horizontal. The coefficient of friction is 0.16. If the crate moves 3.0 m, what is the total work done on the crate?

Guys. So in this problem, I have a bunch of forces that are acting on these crates, and I'm asked to figure out the total work that's done on the crates. So let's go ahead and just draw a quick little sketch with all the forces that are acting. Right? So with this flat surface, my box is like this, and I've got this crate. Their mass is equal to 51 kilograms. Now I have the force. The first force that I'm told is an an applied force of 100 newtons. This is my f a. This is a 100 newtons, and it's directed at 37 degrees above the horizontal. Right? So it's at some angle. Now I also have an mg. So mg acts down. I've got my normal force which acts straight up. Right? The the surface push. And then I've got the force of friction that acts to the left. This is my FK. So these are all the forces and now we wanna figure out the network that is done on the crate by all these forces. So that's gonna be w nets, and remember that there's 2 ways we can figure this out. We can use the sum of all the works. We can just add up the works done by all the forces by using FD cosine data a bunch of times, or you can figure out the net force and then just use FD cosine data once. Either one will get you the right answer. But, in general, if you have all the forces or you can figure out the magnitudes of all these forces, it's probably just best to use the sum of all works. Right? So you're just gonna add up together all the works like this. So really, we just have 4 forces. We're gonna have the work that's done by f a, the applied force, the work that's done by friction, the work that's done by m g, which we know is 0, and then the work that's done by normal, which again is is 0. Your displacements, right, we're told that this box is actually gonna displace 10 or sorry, 3 meters, to the right like this. And so, therefore, your vertical forces actually don't do any work. So we really just have these two forces to figure out right here. So let's go ahead and do that. The work that's done by your applied force is just gonna be that applied force times d cosine theta. So let's take let's take a look here. Your applied force is a 100. Right? We're told that. Your displacement here is 3. Now we just have to figure out the angle that's between those. Well, in our diagram, your applied force is acting at that 37 degree angle, and your displacement acts purely to the right like this. So the angle that we plug into our cosine theta equation is really just the 37 degrees that we've written here. So you're gonna do a 100 times 3 times the cosine of 37. If you go ahead and work this out, you're gonna get a work done by 240 joules, and it's positive. Alright. So now we just all all we have to do is just figure out the work that's done by friction. Now remember that the work that's done by friction is gonna be negative f k times d because your friction is always gonna be going against the direction of motion. So really we just have to figure out what is the apply what is the friction force. Remember, we we can rewrite this equation by saying it's mu k times the normal force times d. Right? So now all we have to do is we just have to figure out this normal force. Is it equal to mg? Well, remember that in these cases, in these kinds of problems where you have forces at 2 dimensional or at 2 dimensional angles like this, and then the normal is actually not gonna be equal to mg. There's a component of this applied force that acts in the vertical direction. So the first thing we have to do is figure out what this normal force is. So to figure out normal, we just have to go back to our sum of all forces in the y axis equals mass times acceleration. Right? This is just our f equals m a in the y axis. Now we just pick our, you know, upward direction to be positive, and then we just add up all the forces. So we have our applied force. There's a component of it that acts in the y axis. When you decompose this vector here, there is a y component of this f a. So we have that plus the normal force and then we have minus mg. And what about the acceleration? Well, in these cases, since the box is only accelerating to the right, the acceleration is equal to 0. So all of these forces have to basically cancel out. So if you solve for our normal force, your normal force is just gonna be mg minus, and then this f a y here, remember, is just f a times the sine of theta. Now we have all of these numbers here. Mg, we have f a, and we have theta. So we can go ahead and figure this out. Your n is just gonna be 51 times 9.8 minus 100 times. And then this is just gonna be sine times, the sine of 37. So if you got to work this out, you're gonna get 440 here. So really what we're gonna do is we're gonna plug in this this 440 back into this equation to figure out the work done by friction. So the work done by friction is just gonna be negative, and this is just gonna be the mu k, which is 0.16 times 440 times the displacement or the distance of 3. You're gonna get 200 or actually rather negative negative 211 joules. So now all we have to do is we just have to add up both of these works together. So we have the work that's done by the applied force, which is 240, plus the work that's done by friction, which is negative 211, and you're just gonna get a total work done that's 29 joules. Alright. So that's really the answer. So if we go back to our answer choices, the answer choice is a. So that's it for this one, guys. Let's move on.

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To pull a 51 kg crate across a rough floor, a worker applies a force of 100 N, directed 37°above the horizontal. The coefficient of friction is 0.16. If the crate moves 3.0 m, what is the total work done on the crate?

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