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Ch 10: Interactions and Potential Energy

Chapter 10, Problem 10

A horizontal spring with spring constant 100 N/m is compressed 20 cm and used to launch a 2.5 kg box across a frictionless, horizontal surface. After the box travels some distance, the surface becomes rough. The coefficient of kinetic friction of the box on the surface is 0.15. Use work and energy to find how far the box slides across the rough surface before stopping.

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Welcome back. Everyone here is the system that we are making observations about. We have a lab bench that is split into two sections here. Now, the first section here is going to be frictionless and the second section is going to have some sort of friction on top of the surface here. Now we have a block that is going to be launched by a spring loaded piston here and here are a couple other things that we are told about this system. We are told that the coefficient of friction in the friction section is 0.2. The block being launched is going to have a mass of 4.2 kg and we are told that the spring constant for the piston here is going to be 150 newtons per meter. Now, we are told that the piston or the spring is deformed to a length of 0.25 m and we are tasked with finding how far the block is going to slide on the friction portion, only not the frictionless portion, just the friction portion here. Well, what let's let's think about this here on a friction and on a frictionless surface with no acceleration. Once the block leaves the piston, it's just going to travel with the same velocity, meaning that it's going to enter the velocity or sorry, enter the friction section with a certain velocity, which we will just label our initial velocity for our sake here. And then it'll stop at some point with a final velocity where the final velocity is just going to be zero because it has come to a stop here. So what can we use in order to figure this out? Well, let's think about the work energy theorem here. We have that net work is equal to a couple of different things. Here, we have the summation of the forces times the distance is covered. That's our desired distance. But we also have that the work is equal to the negative of the change in kinetic energy. Now, we want to find this negative X. So we're going to have to find what our force is and we are going to have to find our change in kinetic energy here. Let's start with our change in kinetic energy. What we can use is the principle of energy conservation. And we can say that in the first part on the frictionless section, we can say that the elastic potential energy is going to be fully converted to kinetic energy. There's no other forces acting on the block except the what you could say, launching power of the piston in the spring. So this conservation is going to hold true. But let's expand this out a little bit. What we have is one half times our spring constant times our initial deformation minus our equilibrium point of deformation squared equal to one half times our mass times our initial velocity squared. Now, this initial velocity is not this one right here. In fact, what I'm gonna do is I'm going to label this V knot and this is the velocity at which the block leaves the piston on the left hand side of the lab bench here. But if you'll notice on the right hand side of this equation, this is just equal to our initial kinetic energy. So what we can do is we can plug in our terms on the left hand side to get our initial kinetic energy here. So let's go ahead and do that. We have one half times. Let's see our spring constant, which is going to be 150 newtons per meter times our deformation here, which we were told was 25 centimeters or 250.25 m minus our equilibrium point which we're just gonna label as zero m squared. And this gives us our initial kinetic energy value which gives us 4.69 Jews. Wonderful. So now since or once the block hits our middle point here where the friction section of the lab bench starts, we are going to have that. Our work energy theorem is going to take place here. But what is the force that is going to be at play here? Well, the only force that will be acting on the block once it enters the friction section because it's already entering with a certain velocity here. The only force that's going to be acting is a friction force that is going to be acting against the motion of the block. Therefore, it is negative. So what is this equal to? Well, this is gonna be the negative of our coefficient of friction times the mass of our block times the acceleration due to gravity or M G can also be known as just the normal force acting on the block here. So now using our work energy theorem here is what we can safely say we have that our force which is negative UK times M G times our delta X or the distance travel buyer block on the friction section is equal to the negative of our change in kinetic energy. So our initial kinetic energy minus our final kinetic energy which just just really distributing the negative sign here. This is just going to be our final minus our initial, we have our initial kinetic energy. But what is our final kinetic energy? Well, we know that our final velocity is going to be zero. And since kinetic energy is just one half times mass times a given velocity, that means our final kinetic energy is going to be zero. So what this gives us on the right hand side of the equation is negative 4.69, which is just the negative of our initial kinetic energy here. So let me scroll down here just a little bit. Here's what we are going to do. I'm going to divide both sides of the equation by negative mu K times M G divided by negative mu K times M G. And you'll see that, that cancels out all the constant terms on the left hand side. So now what we are left with is a formula for our distance traveled which is delta X is equal to negative 4. divided by negative mu K M G. So let's go ahead and plug in all of our values here. We have negative 4.69 divided by negative 0.2 times 4.2 times 9.8. And what this gives us is that the block travels 0. m in the friction section of the lab bench, which corresponds to our final answer. Choice of D. Thank you all so much for watching. I hope this video helped. We will see you all in the next one.
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Textbook Question
A freight company uses a compressed spring to shoot 2.0 kg packages up a 1.0-m-high frictionless ramp into a truck, as FIGURE P10.52 shows. The spring constant is 500 N/m and the spring is compressed 30 cm. a. What is the speed of the package when it reaches the truck?
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Textbook Question
CALC A 10 kg box slides 4.0 m down the frictionless ramp shown in FIGURE CP10.73, then collides with a spring whose spring constant is 250 N/m. a. What is the maximum compression of the spring?
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