### Video Transcript

A body is projected at 14.28 meters per second up the line of greatest slope of a plane inclined at an angle whose tangent is the square root of two over four. If the coefficient of friction between the plane and the body is three times the square root of two over five, what is the maximum distance that the body can travel up the plane? Take ๐ to equal 9.8 meters per second squared.

All right, so letโs say that this is our plane. And if we consider this angle of the plane, weโre told that its tangent is the square root of two over four. This is the ratio of the side opposite the angle to the side adjacent to it. So the base of our plane is four units long, while the height is the square root of two units tall. Weโre further told that a body is projected at 14.28 meters per second up the incline of this plane. The coefficient of friction between the plane and the body is given. We want to figure out how far up the incline the body travels before it comes to rest. Weโll call this distance ๐.

Knowing that the acceleration due to gravity is 9.8 meters per second squared, letโs now clear some space on screen, and weโll start on our solution by recognizing that as our body moves up this plane, both gravity as well as the force of friction will slow it down. Newtonโs second law of motion tells us that the sum of forces acting on a body equal its mass times its acceleration. And as we look at the forces acting on our body, we can find similar information. We note first that thereโs a gravitational force ๐ times ๐ acting on this body, and thereโs also a normal or reaction force acting on it perpendicular to the surface of the plane.

And finally, as our body moves up the incline, a resistive force to the friction pushes the other way. Weโll label this force ๐น. And we now have all three of the forces that act on this body as it moves up the incline. We want to solve for the bodyโs acceleration along this line of steepest slope. Letโs define the direction down the slope to be the positive ๐ฅ-direction. If we then consider all the forces that act along this dimension. The sum of those forces by Newtonโs second law is equal to the mass of our body times its acceleration in the ๐ฅ-direction. Looking at our sketch of the forces acting on our body, we see that the frictional force ๐น acts in what weโve called the positive ๐ฅ-direction and so does a component of the weight force that weโre highlighting here in orange.

To determine this component, weโll want to know the measure of this angle here in our right triangle. If we return to our original sketch of the bottom of our screen, it turns out that this angle here is identical to the angle weโre indicating on our force diagram. To indicate this, we can give them both the same name. Weโll call this angle ๐. The component of the weight force that weโre interested in then is equal to ๐ times ๐ times the sin of ๐. Since this force acts in the same direction as our frictional force, we can add it to our sum of forces in the ๐ฅ-direction acting on our body.

But then we wonder what is the sin of this angle ๐? Returning once more to our original sketch, note that we have a right triangle where the length of the two shorter sides is square root two and four. By the Pythagorean theorem, the length of the hypotenuse โ of this triangle equals the square root of four squared plus the square root of two squared. This is the square root of 16 plus two or the square root of 18. But then we recall that 18 is equal to nine times two. The square root of nine is three. So a simplified expression for our hypotenuse length is three times the square root of two.

Now that we know the relative length of our triangles hypotenuse, we can solve for the sin of ๐. Itโs equal to the side length opposite that angle divided by the hypotenuse length. We see here that the square root of two cancels from numerator and denominator so that the sin of ๐ is simply one-third. Therefore, ๐ times ๐ times the sin of ๐ can be written as ๐ times ๐ times a third. Along with this, itโs possible to write the frictional force ๐น in a different way. In general, for a moving body, the frictional force it experiences is equal to the coefficient of friction multiplied by the reaction force acting on the body.

In our scenario, our reaction force ๐
is equal in magnitude to this ๐ฆ-direction component of the weight force. This component is given by ๐ times ๐ times that cos of the angle ๐. So if this represents the reaction force magnitude ๐
, we just need to multiply this by the coefficient of friction ๐ to arrive at an alternate expression for the frictional force ๐น. And weโre able to do a bit more work on this expression because we can solve for the cos of ๐. Itโs equal to the side length adjacent to the angle ๐, four, divided by the length of our hypotenuse, three times the square root of two.

So, this is our simplified expression for the frictional force ๐น that our body experiences. And if we add this to the weight force it experiences in the ๐ฅ-direction, then that sum is equal to the mass of our body times its acceleration along this dimension. Notice that the mass of our body ๐ appears in every term in this expression. This means if we divide both sides of the equation by mass, then it will cancel out from our entire equation. We can then factor the acceleration due to gravity from both terms on the left, and we can then plug in the given value for ๐ so we can solve for ๐ sub ๐ฅ in terms of the acceleration due to gravity ๐. ๐ sub ๐ฅ equals ๐ times four over three root two times three root two over five plus one-third.

We see that three times the square root of two cancel in this term for numerator and denominator. And then four divided by five plus one divided by three is equal to 17 divided by 15. So, this is the rate at which our body decelerates while it slides up the incline. This constant deceleration is vital information to help us solve for the distance ๐ it takes for our body to come to a rest. Because this body does change speed at a constant rate, its motion can be described by what are called the equations of motion. There are four of these equations in total, but weโll use this particular one. It tells us that the final velocity of a body squared is equal to its initial or original velocity squared plus two times its acceleration multiplied by its displacement.

If we rearrange this equation algebraically to solve for the displacement ๐, we see itโs equal to ๐ฃ sub ๐ squared minus ๐ฃ sub zero squared all over two ๐. Now, in our case, we know the final velocity of our body is zero because weโre finding out how far it takes to come to rest. So, ๐ฃ sub ๐ squared is simply zero. And we know ๐ฃ sub zero. Thatโs given to us as 14.28 meters per second. Therefore, the numerator of our fraction looks like this. And we can now plug in for the acceleration ๐ that we solved for earlier. Itโs equal to ๐ times 17 over 15. And then we can substitute in for ๐, which we know is 9.8 meters per second squared.

Weโre just about ready to calculate ๐, but notice that if we do, weโll get a negative result. This is actually consistent with the sign convention weโve adopted since weโve said motion down the incline is positive. Technically, weโre calculating a displacement which is different from a distance. Our problem statement, though, does ask for a distance rather than a displacement. To make the switch to calculating a distance, we only need to take the absolute value of our answer. When we calculate this fraction that way, we find a result of 9.18. And this distance is measured in units of meters. Our body then travels 9.18 meters up the slope before it comes to a stop.