FIGURE EX10.24 is the potential-energy diagram for a 500 g partic...

Question

FIGURE EX10.24 is the potential-energy diagram for a 500 g particle that is released from rest at A. What are the particle's speeds at B, C, and D?

Accepted Answer

Welcome back everybody. We are making observations about some objects which you are told that has a mass of 800 g or 8000.8 kg. As you can see over here, we are given a graph of its potential energy with respect to its position at different points. And we are tasked with finding one, what is the speed through point P and two? What is the speed through point Q? Let's go over these answer choices real quick before we begin here. So for answer choice, A we have that the speed through point P is 3.16 m per second and zero m per second through point Q. For answer choice B, we have 3.34 m per second through P and two m per second through Q through C or for answer choice C we have 5.12 m per second through P and 10 m per second through Q or for answer choice D, we have 6.184 m per second through P and six m per second through Q. All right. So here's how we're going to do this. We know that by the conservation of energy we have that the total energy of the system is kinetic plus potential energy from this. And looking at just 0.0 right here, we can actually calculate the total energy of the system at 0.0. There is zero kinetic energy and there's six Jews of potential energy giving us a total energy of six jewels for the entire system. I'm gonna revisit this initial equation over here. What I'm gonna do is I'm gonna subtract potential energy from both sides. What this now gives us is that kinetic energy is equal to the total amount of energy for the system minus the potential energy at a given point. But remember what we're trying to find is speed. So we seem to somehow relate this formula to another formula. We also know however, that kinetic energy is just equal to one half times the mass of the object times the velocity squared at that point. So what I'm gonna do is I'm gonna sub in this value for kinetic energy into this equation right here. What this gives us is that one half M V squared is equal to the total amount of energy minus the potential energy of the system. Now, we want to isolate this velocity variable right here. So here's what I'm going to do. I'm gonna start off by multiplying both sides by two over M two over M. And you'll see that this cancels out all of these values on the left hand side now to get rid of this exponent on the velocity variable, then what we are going to do is just take the square root of both sides making this variable disappear. Leaving us with the equation that velocity at a given point is equal to the square root of two times our total energy of our system minus the potential energy at a point all divided by the mass of the object. Now that we have that formula, we are ready to go ahead and find the velocity at each of our given points. So let's go ahead and start with part one here. We have that the velocity at P is going to be equal to the square root of two times six, which is the total energy of the system minus our potential energy at point P which looks to be two divided by .8. And when you plug this into your calculator, you get 3.16 m/s. Great. So now let's go ahead and move on to part two here. Same process for velocity through point Q velocity through point Q is going to be equal to two times six minus the potential energy at that point which is six all divided by 60.8, which as you can see, we get zero m per second. So now we have found the velocity through point P and the velocity through point Q which corresponds to our final answer. Choice of a 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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FIGURE EX10.24 is the potential-energy diagram for a 500 g particle that is released from rest at A. What are the particle's speeds at B, C, and D?

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