A skier is gliding along at 3.0 m/s on horizontal, frictionless snow. He suddenly starts down a 10 degree incline. His speed at the bottom is 15 m/s. (a) What is the length of the incline?

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Hey, everyone in this problem, we have a box that's initially moving on a flat and smooth surface with a speed of one m per second that is projected downward along a frictionless inclined plane. The plane makes an angle of five degrees with the horizontal. And at the end of the incline, the boxes speed is increased to five m per second. We're asked to calculate the distance traveled by this box along that incline. We're given four answer choices. Option A 2.34 m, option B 10.3 m, option C 14 m and option D 56. m. So we're gonna start by drawing our incline. So we have our incline and it makes an angle of five degrees with the horizontal. Ok. And our box is gonna start at the top of this incline and it's gonna have a velocity or speed of one m per second. After some time, it's gonna reach the bottom of this incline. It's gonna still be moving and now it's gonna be moving with a speed which we're gonna call V fa final speed of five m per second. Now we're gonna take to the right and parallel to that incline as our positive direction. OK. And so that incline is gonna be a tilted axis for us, it's gonna be our X axis. And so now all of this motion is occurring in the X direction, we only have one direction of motion to worry about. All right, let's think about in this X direction. What variables we have now, we've already written our two speeds. So we have the initial speed V knot is one m per second. The final speed V F is five m per second. And, and when we're talking about velocities, these are gonna be positive velocities because they're happening in our positive direction. Now, delta X that horizontal distance traveled, that's what we're trying to find. OK. That's gonna be that distance traveled along the incline. So delta X is what we're looking for. Now, in terms of acceleration, we know that we're gonna have gravity acting here. OK? Because this box is at a particular height. So we have gravity acting straight down vertically and we can project that gravity onto our tilted X axis K onto that incline. And the horizontal acceleration of this block is gonna be pointing to the right. We can see that we have a triangle here. The hypotenuse is G and the angle with the perpendicular to the incline is gonna be five degrees. All right. So what that means is our acceleration that we're looking for that horizontal acceleration. A is gonna be related to this angle through sign because it's on the opposite side. OK. So this acceleration is gonna be signed of five degrees multiplied by the acceleration due to gravity 9. m per second squared. All right. So we have three known values. We have one thing we want to find. So we're gonna choose a kinematic equation that has these four variables where that we've written it and we're gonna substitute in what we know and we're gonna solve for delta X. And you'll notice we haven't written t here. And that's because we don't have information about T and we're not looking for information about T. So we can leave that out. We get the kinematic equation V squared is equal to V not squared plus two A delta X substituting in our values V F five m per second. And that's all gonna be squared. This is gonna equal V naught squared, one m per second squared plus two, multiplied by the acceleration stine of five degrees multiplied by 9.8 m per second squared, multiplied by delta X that horizontal distance or that distance along the incline that we're looking for. Now, let's move our one m per second squared to the left hand side as we work to isolate delta X. And we're gonna simplify as well five m per second, all squared that's gonna give us 25 m squared per second squared. And we're gonna subtract this one m per second, all squared, it's gonna be minus one m squared per second squared. And this is equal to, we have two multiplied by 9.8 m per second squared. This gives us 19.6 m per second squared. We are still multiplying by sine of five degrees and by delta X. Now isolating for delta X, we wanna divide by everything else. On the right hand side, we get that delta X is gonna be equal to what we originally had. On the left hand side, we had 25 m squared per second squared minus one m squared per second squared. That's gonna be 24 m squared per second squared. And all of this is gonna be divided by what was on the right hand side with delta X. OK? So 19.6 m per second squared multiplied by sine of five degrees, right? In terms of our units, we have meters squared per second squared, divided by meters per second squared. That's gonna leave us with a unit of meters, which is exactly what we want when we're talking about a distance. And if we work this out on our calculator, we're gonna get 14. m. And that is that distance traveled along the incline. If we go up to our answer choices and compare what we found, and we can see that these are rounded to three significant digits. And if we round what we found we see that the correct answer is option C 14 m. Thanks everyone for watching. I hope this video helped see you in the next one.

A skier is gliding along at 3.0 m/s on horizontal, frictionless snow. He suddenly starts down a 10 degree incline. His speed at the bottom is 15 m/s. (a) What is the length of the incline?

Verified Solution

Key Concepts

Conservation of Energy

Kinematic Equations

Inclined Plane Dynamics