Pearson+ LogoPearson+ Logo
Start typing, then use the up and down arrows to select an option from the list.

Finding Speed of a Crate Using Work-Energy Theorem

Patrick Ford
Was this helpful?
Hey guys, when this problem I have a crate that sliding down an inclined plane and it has some forces acting on it and we have to calculate some works. So let's go ahead and just draw a quick little sketch here. So I've got this incline plane like this. Going to draw a nice and big here. I've got my box that is seven kg and I've got some forces acting on it. So let's go ahead and label all the forces. So I've got MG. I know that's that's acting. I've got the normal force, we've got a component of MG that's acting down the ramp, this is MG. X. And this is going to be my MG. Why? And then I also have the force of friction as the block slides down the ramp, friction wants to oppose that and want to go to play up the ramp like this. So I got my friction force. So now what I want to do is I want to calculate the work that's done by gravity if the crate slides 2.5 m down the ramp. So basically that's this distance like this d equals 2.5 here in which it's going to slide down the ramp so it's going to end up over here somewhere. So let's go ahead and do that in part a and part. I want to calculate the work that is done by gravity. So it's W. M. G. Now remember that on an inclined plane W. M. G. Is the same thing as W. M. G. X. And so I can just say this is just gonna be MG times the sine of theTA X. This angle here is the 26 degrees. So this is day to exit the sequel 26 degrees. And um it's gonna be MG. Sine theta X times the distance. So what I have to do is first I have to pick a direction of positive. Now what happens is my MG. X. Is gonna actually point in the direction of my displacement. And so what I'm gonna do, I'm just going to have my positive direction be this way. So what this means here is that your W. M. G. Is just going to be, this is gonna be positive and so therefore I'm just gonna start plugging stuff in. So I've got seven times 9.8 Times the sine of 26°.. And then I've got the distance here, which is 2.5. If you go ahead and work this out what you're gonna get, you're gonna get 75.22 jewels. Makes sense that this is positive because the block is going to start going faster because of this work here. All right, so what about part B. And part B. Now? I want to calculate the work that's done by friction. So, remember that when we calculate the work done by friction, it's always going to be negative F. K. D. So, this is just gonna be once I expand this out negative mu kinetic times the normal times distance. So really the trouble is that I wanted to figure out this normal force on an inclined plane. Now, on an inclined plane, if you're only if you're only forces act horizontally, then your normal force is going to equal M. G. Y. Which is just gonna be MG time to the co sign of theta. X. So, what I can do here is I can see that the work that's done by friction is gonna be negative mu K. And then the normal force really just becomes MG co signed tha tha X. And then I have D here. So, I looked through all my variables. I have Mieux que I have the mass gravity, I have the angle and the distance. So I can calculate the work done by friction. So my work done by friction is gonna be negative Times 9.8 times. Oh, I'm sorry, the mass, the mass is seven massive seven. Now we have 9.8, so now we have the co sign of the angle which is 26 degrees and this times the distance, which is 2.5. So you go ahead and work this out. What you're gonna get is 55.5 except it's gonna be negative. So it's negative. 55.5 jules makes sense that it's negative because as the block goes down, grab so the first force of friction is going to want to remove energy from the block and slow it down so it's negative. All right. So now, in this last part here, we have to find the speed of the crate if it starts from rest. So the idea here is that the initial velocity is going to be zero here and then when it gets down to this point is going to have some final velocity. And I want to figure out what that is. So I just calculated a bunch of works and I can relate them to speeds by using the work energy theorem. Remember that the network is just equal to the change in the kinetic energy. So the work that we just calculated the work that we just calculated 75.2 and negative 55.5. We can combine them to find the network. So our network is really just gonna be our MG plus the work done by friction. Plus the work done by normal. Now, in this case the normal doesn't do any work right? We know that. So this is equal to the change in the kinetic energy. So there's really just K final minus K. Initial. However, because the initial speed is equal to zero, there is no initial kinetic energy and all of it just goes to the kinetic energy final. All right. So that's all we got to do. So really we just have this 75.2 that we just plug in plus negative 55.5. And this equals the kinetic energy. Final. This is gonna be one half. M. V final squared. So we want to find this V final here. We're just gonna have to solve everything else. So agenda getting When you subtract these two things is you get 19.7 this equals one half and then times seven times the final squared we rewrite that again. So this is seven here. So when I move everything over to the other side, what I'm gonna get is 19.7 times two times two divided by the mass of seven. And this equals DV final squared. So when you go ahead and plug this in and take the square roots, what you're gonna get is that your V final is equal to 2. m per second and that's your final answer. All right guys. So that's it for this one. Let me know if you have any questions.