An 8.00-kg block of ice, released from rest at the top of a 1.50-...

Question

An 8.00-kg block of ice, released from rest at the top of a 1.50-m-long frictionless ramp, slides downhill, reaching a speed of 2.50 m/s at the bottom. (b) What would be the speed of the ice at the bottom if the motion were opposed by a constant friction force of 10.0 N parallel to the surface of the ramp?

Accepted Answer

Hey, everyone in this problem, we have an experiment where so holds a stationary wooden block of mass 5.8 kg at the top of a 1.8 m long frictionless incline. OK. So let's just start drawing this out as we go. So we have this incline, you know that it is 1.8 m long. We have a block starting at the top with the mass of 5.8 kg. Now, the student is gonna release that block and measure its speed at the bottom of the ramp to be 3.3 m per second. And we're gonna call that the final speed V and that's gonna be 3.3 m per second. Now, the student is gonna repeat the experiment and we're asked to determine the speed of the block at the bottom of the incline. If a constant 4.6 Newton friction force parallel to the incline surface opposes the motion. OK. So this second repeat experiment is going to be with friction. All right. In our incline, we're just gonna draw, make an angle of theta with the horizon. Now we have four answer choices here all in meters per second. Option A 5.69 option B 5.94 option C 2.83 and option D 1.69. Now, what do we need? Hey, if we wanna think about this repeat experiment, we wanna figure out what the speed at the bottom of the incline is. We're gonna need to know this value of data and we have to know this value of data in order to figure that out. So what we're gonna do is work with the first experiment where we know the speed at the bottom of the incline in order to find theta. And then we can use that to find the speed in that second experiment. So let's start with a free body diagram. We're gonna wanna use our forces here. So we have our block, we know that there's a normal force acting perpendicularly upwards from that surface of our incline. We have the force of gravity acting straight down and we're gonna break that into two components. We're gonna call them FGX. OK. That lies parallel to the incline. And then we have FGY lies perpendicular. Yes, we're just breaking them up in that direction and our block will be accelerating to the right since it's moving downwards on the incline. Now, if we think about our forces, OK, we have that the sum of the forces in what we're calling the X direction. OK. That direction parallel to our incline is gonna be equal to the mass multiplied by the acceleration of the block. OK. That's Newton's second law. Now, the forces we have acting in this direction, all we have is the X component of the force of gravity. So what we have here is F GX, it's going to be equal to the mass um multiplied by the acceleration A FG. OK. The force of gravity is given by M multiplied by G. In this case, we want the X component. And so we're gonna have mg multiplied by sine of theta is equal to the mass multiplied by the acceleration. Now, remember what we're trying to do. We're trying to find theta so that we can use that in this second experiment to find the speed. OK? We can see that in our equation, the mass M terms divided we're left with G stein of the is equal to A. OK. So we have the, in our equation. Now we know G but we don't know the acceleration A and so we can find the acceleration A then we'll be able to substitute it into this equation and find the OK. So let's think about the motion along the incline. We know that we have an initial speed V nat of 0 m per second. It's a stationary block that the student releases. So the initial speed is zero, we know that the final speed or the final velocity in this case is positive 3.3 m per second. OK. We're taking to the right to VR positive direction delta X, the displacement in this X direction which were um naming us the direction parallel to our incline is 1.8 m. We don't know the acceleration and we don't know the time T now we have three known values. We want to find the acceleration to use in our equation one which is the equation G sine theta is equal to A. So we're gonna use our kinematic equation with VNVF delta X, the three known values as well as a, the value we're looking for. And that equation is gonna be the following VF squared is equal to V not squared plus two A delta X. So we get 3.3 m per second, all square is equal to zero squared is just zero. So I'm gonna leave that we have two multiplied by the acceleration. A multiplied by 1.8 m. If we divide both sides by two, multiplied by 1.8 m, what we get is that the acceleration is going to be equal to 3.025 m per second squared. So this is the acceleration of the block when we don't have any friction going back to equation one, we can now find the angle theta that our incline makes with the horizontal substituting in our values 9.8 m per second squared. Multiplied by sine of the is equal to 3.025 m per second squared. OK. So the we're gonna do is divide both sides by 9.8 m per second squared. Take the inverse sign and we get that theta is sine inverse 3.025 divided by 9.8. And you can see that I've divided up the units, we have meters per second squared, divided by meters per second squared. And so we end up with just a number here, no units inside of that um inverse trig function. And this gives us a value of theta of approximately 17.98 degrees. OK. So we're gonna put a red box around that. So we don't lose track of it, but that is the angle that our incline makes with the horizontal. Now, let's go look at our picture again and remember what we're trying to find in the end here, we wanna find the speed of the block at the bottom of the incline. In the case where we have a friction, we know this angle theta now, so now we can work on the second part of this problem actually solving this repeat experiment. OK. So let's do that. And what we're gonna do is draw a free body diagram for the case of friction and we're gonna switch over to blue. OK? All of our writing in black is going to be the case where there is no friction. All of our writing in blue will be the case where we have friction. OK. Just so we can distinguish between those two. So we have our block again, gliding down the incline, we still have a normal force perpendicular to the incline. We have our force of gravity pointing downwards which can be broken up into an X component that's parallel to the incline in a Y direction that is perpendicular to the incline. So, so far, this free body diagram is the exact same as the first one we drew. But now we have friction and that is going to point to the left, OK, along that incline because it's going to oppose the motion. We have our friction force f pointing to the left opposing the motion. Now we might be thinking, well, let's just use our kinematic equations like we did. In the first case, we don't need to worry about our free body diagram and our forces. So let's write out our variables and see what we have. And we have that var it's going to be equal to 0 m per second again, starting from rest. The at the final speed is what we are looking for. Delta X is still 1.8 m and the incline has changed, it still has the same length. The block is still traveling the same distance. We don't have the acceleration in this case. And remember it's not going to be the same because we have that force of friction and we don't know the time it takes. So right now, we only have two known values which is not enough to go ahead and solve for VF that final speed at the bottom of the incline that we want. So we do need to go back and use our forces just like we did. In the first case, we can use them to find our acceleration and then come back to our kinematic equations. So again, we're gonna look at the sum of the forces in the X direction that's gonna be equal to the mass multiplied by the acceleration. In this case, our sum of forces in the X direction, we have the force of gravity, the X component in the positive direction and we have the force of friction in the negative direction. So FGX minus F gonna be equal to M multiplied by A just like before the X component of our force of gravity. M multiplied by G multiplied by sin of the, that's gonna be minus our force of friction that's equal to M multiplied by A. Now we can go ahead substituting our values. And so for a um mass 5.8 kg multiplied by the acceleration due to gravity 9.8 m per second squared multiplied by sin of the oops sorry we substitute in our values. We have s of the. Now we can do two things here. We can go sign of the angle we found 17.98 degrees. What we can also do is notice that we had sin of theta equal to 3000 and or 3.025 divided by 9.8. OK? So we can just use that value. We avoid any round off error. If you didn't realize that and you just use the data value, you found that's OK. We're gonna get the same answer. We are gonna use that exact value. OK? So sign of data, we're gonna use 3.025 divided by 9.8 and no units there because they divided it. All right. Now we're subtracting our force of friction, which we're told in the problem is 4.6 newtons and this is all equal to the mass 5.8 kg multiplied by the acceleration A. We're gonna work out the left hand side divide by 5.8 kg. And we get that the acceleration A is going to be equal to two 0.2319 m per second squared approximately. And this is a good time to stop and make sure this makes sense, make sure we're not doing anything wrong. And this does kind of make sense. OK. Our acceleration in the first case, let's go back to it. We found to be 3.025 m per second squared. Now we've added friction and we found an acceleration that is smaller than that initial acceleration. OK? So that makes sense. We have friction opposing the motions the acceleration is less. OK. So we're on the right track, we have one more step to do we have our acceleration. Now, we can use that in our kinematic equations. OK. So we have the acceleration as 2.2319 m per second squared. And once again, we're gonna choose the equation that has VNO delta X and A are three known values and VF are unknown value that we are interested in. That's the same equation as before. VF squared is equal to V not squared plus two A delta X. We get that VF squared is equal to while V not squared is just going to be zero. So we get two multiplied by 2.2 3 1 9 m per second squared multiplied by 1.8 m. If we take the square root, we get that our final speed. VF is going to be two 0.8346 meters per second and that is the speed at the bottom of the incline four hour. OK. So we had a lot of steps there but we just needed to work through them one at a time. OK. So we started by looking at our forces, OK? We wanted to find the theta value the angle that this incline makes. In order to do that, we needed to calculate the acceleration in the case where we had no friction. OK. So we used our kinematic equations to do that that allowed us to find the angle of the incline. Once we had the angle of the incline, we could find the acceleration of our block when we had friction. And then once again, use kinematic equations or U AM equations to find that final speed at the bottom of the incline, comparing this to our answer choices and rounding to three significant digits. We can see that the correct answer is option C 2.83 m per second. Thanks everyone for watching. I hope this video helped see you in the next one.

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