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Ch 10: Interactions and Potential Energy

Chapter 10, Problem 10

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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Welcome back, everyone. We are making observations about the following system here. We are told that students are using a spring to fire blocks here. Now, we are told that once the block leaves the spring, it is going to travel up an incline here. So let me actually go ahead and draw my incline out here. Go ahead and give it a vertical and a horizontal. We are told that the incline is 1.5 m high and then once the block travels up the incline, it will go onto the bench here. Now, we are told a couple other things about the system. Of course, like we said, the final height is going to be 1.5 m. Now, the mass of each of the blocks is 500 g or 5000. kg. Now, the spring constant is going to be 650 newtons per meter. And we are told that the spring is compressed by centimeters or 500.5 m. And we are tasked with finding what is going to be the final velocity once the block reaches the top of the incline and slides onto the bench. So what is going to be our final velocity. So what are we going to use here? In order to relate all these terms, we're dealing with springs, we're dealing with inclines and height and whatnot. The best thing to do here is to use the principle of energy conservation here. And here's what it says we have that our final kinetic energy plus our final gravitational potential energy plus our final spring potential energy plus the change in thermal energy is equal to our initial kinetic energy plus our initial gravitational potential energy plus our initial spring potential energy plus any work done by external forces. Here, we have uh formulas for each of these terms. So let's expand out this equation. What we get is one half times our mass times our final velocity squared plus M G times our final height plus one half times our spring constant times our final deformation of the spring squared plus that change in thermal energy is equal to one half M V I squared plus M G. Our initial height plus one half times our spring constant times our initial deformation squared plus the work done by those external forces here. What do we know what we're told that the incline and the bench are both frictionless. This means that there is going to be no change in thermal energy there. Well, there's gonna be no conversion from or to thermal energy in the first place. So that term is going to be zero the block is initially going to start out at rest. So we can say that its initial velocity is going to be zero m per second. There are no external forces working on the block. So that means the work by external forces is going to be zero. And if we consider the ground level or the base of the incline, the starting position, then we can just say that our initial height is zero m as well. So what does this exactly leave us? Well, let's plug in the terms we know into these equations, including all of these terms that are zero. What we get is we get one half times mass times our desired final velocity squared plus mass times gravity times our final height plus zero plus zero is equal to zero plus zero plus one half times our spring constant times our initial deformation here. Oh Yes. One thing I forgot to mention our final deformation since the spring is just going to have returned back to its equilibrium point is simply going to be zero as well. Hence, uh let's see here. Hence this zero right here in our equation. All right, getting back to it. On the right hand side of our equation, we have one half k times our initial deformation squared plus zero. So here's what I'm gonna do. We want to isolate this final velocity term right here. So I'm going to subtract our final gravitational potential energy term from both sides of our equation. And on the left hand side of the equation, it cancels out. And then I will multiply both sides of the equation by two over our mass two over our mass. And you'll see on the left hand side of the equation that cancels out all the constant terms. So now let me scroll down here just a little bit. So we have some more room. All right. So what this gives us is that our final velocity squared is equal to our spring constant times our original deformation squared minus two times our mass times our acceleration due to gravity times our final height all divided by the mass. And finally, to isolate the final velocity term fully, we're just going to take the square root of both sides of our equation. Therefore, getting rid of that second power. Wonderful. So now we are ready to go ahead and plug in all of our terms to find the final velocity here. So what does this give us? We have 650 times 6500. squared minus two times 20. times 9.8 times 1.5, all divided by 0.5. And when we plug this into our calculator, we get that the final velocity of the block once it reaches the top of the incline and 17.2 m per second, which corresponds to our final answer. Choice of C Thank you all so much for watching. I hope this video helped. We will see you all in the next one.
Related Practice
Textbook Question
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?

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Textbook Question
In FIGURE EX10.28, what is the maximum speed a 200 g particle could have at x = 2.0 m and never reach x = 6.0 m?

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Textbook Question
A system in which only one particle moves has the potential energy shown in FIGURE EX10.31. What is the x-component of the force on the particle at x = 5, 15, and 25 cm?

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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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Textbook Question
The elastic energy stored in your tendons can contribute up to 35% of your energy needs when running. Sports scientists find that (on average) the knee extensor tendons in sprinters stretch 41 mm while those of nonathletes stretch only 33 mm. The spring constant of the tendon is the same for both groups, 33 N/mm. What is the difference in maximum stored energy between the sprinters and the nonathletes?
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Textbook Question
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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