The spring shown in FIGURE P10.54 is compressed 50 cm and used to launch a 100 kg physics student. The track is frictionless until it starts up the incline. The student's coefficient of kinetic friction on the 30° incline is 0.15. a. What is the student's speed just after losing contact with the spring?

Video transcript

Hey, everyone. So this is a conservation of energy problem. Let's see what they're asking us. A science teacher pushes a 0.75 kg cube against an ideal spring with a spring constant of 300 newtons per meter. The spring is compressed to a distance of 5cm. The teacher releases the Cube after moving over the smooth horizontal part of the setup shown in the figure below the cube slides down over a rough ramp Making an angle of 15° with the horizontal, the coefficient of friction, the coefficient of kinetic friction between the Cube and the ramp is .25, determine the speed of the Cube just after leaving contact with the spring. And our multiple choice answers here are a one m per second. B 1.2 m per second. C 4.5 m per second or D 20 m per second. So the first thing we can do here is recall that our conservation of energy equation means that our potential energy. So we'll call it the potential energy of the spring plus the kinetic energy when the uh not of the spring. But when it is compressed, when the block is compressed against the spring is equal to the potential energy right before it reaches the ramp plus the kinetic energy right before it reaches the ramp. And so we can recall that our potential energy is for a spring is given by one half K multiplied by delta X squared. Our kinetic energy is one half or kinetic energy is one half M V squared. And that's, well, that will be the same for the other side of the equation. So one half K delta X squared Plus 1/2 and the squared. Now for this first scenario, the first part of this problem, we have the um cube compressed To a distance of five cm. The teacher releases the cube, which means that when the cube is compressed, there is no speed. So our, our V I here the initial is going to be zero. So this term goes to zero. So this part of the equation simplifies to one half K and delta X We know is that Cube is compressed five cm. So I'm going to put that in um standard units of meters. And so that's gonna be negative 0.5 m squared. So I'm just gonna come down here and make sure it's clear how I got that. So delta X initial is equal to the final position, which is that spring equilibrium position that we are going to give, We are going to assign that value zero m and then its initial position. So final position is zero m, initial position is that compressed five centimeters. So zero m minus 0. m. And that is how we get that negative 0.05 m for Delta X. OK. So that is the left half of this equation. Let's look at the, so that's the initial half, let's look at the final half. So just after we, this block has left contact with the spring, there is no potential spring energy. So this term goes to zero and then we are left with one half M V F squared and V F is what we are solving for. So here the haves canceled, just make it a little bit easier. We know our um spring constant And the problem was given to us as 300 newtons per meter. So I'm just gonna write that down here. So we have K equals 300 newtons per meter. We've already. So for delta X initial, The mass was given in the problem as .75 kg. And then we're solving four V E F so we can rearrange this equation to solve to isolate V F. And so V F is going to be the square root of K multiplied by delta X I squared, which we established as 0.5 m squared, all of that divided by the mass. So we'll plug in the remaining information K 300 newtons per meter multiplied by negative 0.5 m squared, All divided by 0.75 kg. We plug that into our calculator and we get 1.0 m/s. So that is the answer and that aligns with answer choice. A that's all we have for this one. We'll see you in the next video.