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Problem 112 Hard Difficulty

Consult Multiple-Concept Example 10 for insight into solving this type of problem. A box is sliding up an incline that makes an angle of $15.0^{\circ}$ with respect to the horizontal. The coefficient of kinetic friction betwecen the box and the surface of the incline is $0.180 .$ The initial speed of the box at the bottom of the incline is 1.50 $\mathrm{m} / \mathrm{s}$ . How far does the box travel along the incline before coming to rest?

Answer

0.27 $\mathrm{m}$

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Video Transcript

in these questions begin by calculating the acceleration off this box for then calculating how far it's lives up before coming to rest. To calculate the acceleration, we have to use Newton's second law. For that, I would choose the following reference frame. My white axis will be these access making 90 degrees. We've these loop and my X axis will be parallel to his look, pointing to the direction of movement. Then, using these reference frame, we apply in Utah's second law. The exact says. It tells us the following the Net force that direction is equal to the mass off the box times its acceleration in X direction, and we are trying to calculate precisely its acceleration in the X direction. Then I noticed that the Net force in that direction is composed by cheer forces, the frictional force that points to the negative direction and the X component off the weight force. Which ouster points to the narrative X direction. The components off the weight force are as follows. This is the Y component off the weight force, and this is the axe component off the weight force. Then Newton's second law tells us that minus frictional force. Minors wait for us. Component X is equals to the mass times acceleration off the box. Then the acceleration is given by minus the frictional force plus the weight force component X divided by the mass off the box. But the frictional force in question is the kinetic frictional force, which is given by the kinetic Gino Coefficient times the normal force so we can write it as acceleration equals two minus muche eight times the normal force plus wait Force component acts divided by the mass. No, we can We have to calculate what is the X component off the weight force? For that, we can use this triangle. So this triangle is like this. This is the high part in use. We have one side. We have the other side. Here we have a 90 degree angle. The high party news is the magnitude of the weight force. These is the Y component. And is this tax component? Let us use these angle. Ask Tita then, by calculating the sign off. Peter, we can better mind what is the relation between the X component and the magnitude off the weight force? Because the sign off Peter is give him by the opposite side w X, divided by the high Pontin use, then the value of X is equal stewed w times this sign off dicta, then the acceleration he's given by minus UK times the normal force plus W times this sign off, divided by the Mass off the box. And then remember that the weight force is given by the Mass off the box times acceleration of riveting. So we got an expression for the acceleration that is as follows genetic, additional coefficients times the normal force plus m times g times this sign off detail divided by the mass off the box. And now we have to calculate what is the normal force for that? We have to apply Newton's second law in the Y direction. Actually, we don't need to do so because it's very easy to see that the normal force he's been exactly compensated by the Y component off the weight force. But in any case, being more regulars, we can use Newton's second law, which tells us the following the Net force in the direction is because the mass off the box times its acceleration in the direction the box is only moving on the X direction. So its velocity in the wind direction is a question zero. And this velocity is not going to change their afford. The acceleration is it close to zero. Then the net force in the Y direction is it goes to zero. But the net force in the Y direction is composed by true forces the net force which points in the positive direction and the Y component off the weight force which points the negative direction. So the normal minor is the way to component. Why is the question zero then? As I said, the normal force is equal to the white component off the weight force. And then we can use this triangle again to complete the relation between the Y component off the weight horse and a magnitude of the weight force. We used the co sign for that because a co sign off Tita is given by the address inside the way. Why divided by the hypotenuse w. Then you wait force why component is given by the weight times. Niccum Sign off Tita which means that the normal force his equals to do it. Go sign off detail. Then we can go back expression for the acceleration and substitute the normal force by these expression to get that the acceleration in that direction is given by minus from UK times the weight which is m times, g times the co sign off Tita Plus I am g times the sign off detail and these wall expression is divided by the mass off the box. Notice that there is a common factor of em in the numerator and a factor off. I am in the denominator so you can simplify it and then we get the final expression for the acceleration next direction. It's even by minus, um UK Times to co sign off Dita plus sign off Teeter on these wall thing is multiply it by the acceleration of gravity. So this is expression for the acceleration in the X direction. Now, Before evaluating this expression, let me organize my board. Okay, Now you're in condition off evaluating the acceleration it's given by minus 0.180 times the co sign off detail. But what is Teeter? Teeter is this angle here? Inside is triangle, so these is teeter. But what is the value off detail? It's not difficult to see that teeter is because to 15 degrees. If you have a hard time to see that, just look at what I'm doing. We can extend the weight forced down here so that we form Iraq. Tongo varies that has a 90 degrees angle inside it, Then 15 degrees plus X shoots out should add up to 90 degrees. It means that this angle of here is an angle off 90 minus 15 degrees, so it's an angle off 75 degrees. On the other hand, the angle between this loop on the Y component off the weight force is also an angle off 90 degrees. Therefore 75 plus teeter must be 90 then teeter is equals to 15 degrees. As I said so now, going back to the violation, we have the curse sign off lifting, plus the sign off 15 and this is multiply it by 9.8. In case you forgot, the acceleration of gravity near the surface off the earth is approximately 9.8 meters per second squared and then these results in acceleration off approximately minus 4.24 zero meters per second squared. So these is the acceleration off the box next direction. Now we are able to evaluate how far the box travels before coming to rest. We have two regions. Equation tells us that the final velocity squared is equal to the initial velocity squared, plus true times the acceleration, times, displacement, the final velocity When the box comes. Rest zero. The initial velocity of the box is 1.5 on. We have plus two times the acceleration off miners four points 24 times they displacement, which we want to calculate. Then we have to solve this equation for the displacement, and it goes as follows. Two times 4.24 times the displacement use 1.5 squared, then 8.48 times the displacement, just 1.5 squared so that the displacement is 1.5 squared, divided by 8.48 which is approximately 0.265 meters. So this is home for the box travels before coming to rest