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I) How much tension must a rope withstand if it is used to accelerate a 1210-kg car horizontally along a frictionless surface at 1.20 m/s$^2$ ?
(I) What is the weight of a 68-kg astronaut ($a$) on Earth, ($b$) on the Moon ($g =1.7 m/s^2$) ($c$) on Mars ($g = 3.7 \,m/s^2$) ($d$) in outer space traveling with constant velocity?
(II) A person has a reasonable chance of surviving an automobile crash if the deceleration is no more than 30 $g$'s. Calculate the force on a 65-kg person accelerating at this rate.What distance is traveled if brought to rest at this rate from 95 km/h?
(II) According to a simplified model of a mammalian heart, at each pulse approximately 20 $g$ of blood is accelerated from 0.25 m/s to 0.35 m/s during a period of 0.10 s. What is the magnitude of the force exerted by the heart muscle?
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in this video, we'll begin looking at the potential energy associated with charges. But before we do, it's helpful to review some concepts for mechanics, and we'll start with work. Which will remember is some applied force times some displacement. But of course, force and displacement are vectors. So say, for example, we have some uniformed force. It's to the right in every direction, and then we move from some point a to some point B. Well, then we have to break up this path and calculate the little bits of work done along every bit of it. Which gives us F d L. We take the dot product because at any particular point, se de l here is pointed in that direction. Then f dot de l will give us the component of work aligned with the force, which is all of the work done. Then we can integrate this from a to B and find the total work done. Of course, that's not the only path Weaken take. We'll call this instead l one and we'll call this work one and integrate overall one. We could also go say on that path from A to B, in which case. When we integrate over work, too, we have have to integrate over the second path. The important thing here is that our total integral is work one and work to are equal. We have a conservative force. In other words, our path doesn't matter. It just matters that where we start and where we end. As with potential energy for gravity. If we started some height B and some height A, we can fall straight and the work done by gravity is one value. We'll call it W G one or say there's some mountain in this direction and we fall down the face of that mountain and still end up at a height. A Well, w g two is the work done by gravity, and it's still equal to double G one. So with conservative forces, path doesn't matter Onley where we start and where we end up. So that's the first thing we want to remember. Second is how work is related to potential energy, and we'll recall that work is simply the negative change in potential energy. In other words, the negative of you final minus you initial. So if we have an increase in potential energy. We've done negative work because this number is positive. Then we have a negative sign. If we have a decrease in potential energy, we've done positive work. And finally, if our forces conservative, the total energy or change in energy when we do work is zero. In other words, work we can define as the change in kinetic energy. But we know also that it's the negative change, potential energy. And so if we rearrange this, we see that our total energy does not change. What we have initially is what we end up fire at the end. And, of course, this is Onley for conservative forces. Now let's consider the potential energy associated with the parallel plate capacitor, and we'll recall that a capacitor it's merely two sheets of equal and opposite charge, and the charge will have uniformly distributed over these sheets. Well, we'll call for a previous video that as long as the sheets are very big and the distance between them very small in the electric field outside of the sheets will simply be zero, and between them will be constant and magnitude and moving from the positive sheet to the negative sheet. And so we can write that R E is some negative e X negativity in the extraction and, of course, is simply the magnitude of this field and negative because it's moving in the negative X direction. That means our force when we put a charge in there, say charge Q. It's simply negative. Q E x So let's place that charge. Q. Here at point A. We're gonna let this force move it to Point B. Well, what's the work done? Work Recall. It's the integral of F d. L and Al de l is going from A to B, but our force and our direction are parallel, so we can just take the full product and not just the dot product. How we get f d. L from A to B, which one we solve. We'll get negative que e b plus q e A. But of course, work is also equal to the negative change in potential energy, which I will write like this because we can see if we can relate the terms in each of these to develop a relationship or expression for potential energy. And what seems to be obvious is that our potential energy, which will depend on X. It's simply going to be Q E X. So does this make sense? As we move from a to B? What kind of work are we doing? Well, we're moving in the direction of the Applied Force, so we expect the work done by that force to be positive, and we'll see here. We can rewrite this as q e times a difference at A and B and since a is greater than be, this is greater than zero. So we've done positive work that makes sense. Our potential energy or the change of potential energy is cute times the difference of B minus A and this is less than zero. So we've lost potential, and that makes sense. Positive work means we are using that potential to turn into kinetic energy or some other form of energy. And negative work would mean that we're adding potential energy into the system, just as we do with gravity. And so these numbers make sense. And so here we have a solid expression for the potential energy associate ID between two parallel plates of equal and opposite charge. Now the next question might be, well, where's potential? Zero and this is exactly like with gravity with gravity, we can say the gravitational potential is zero at sea level or at the top of Mount Everest are at the bottom of Death Valley. It's all relative, and the numbers will work out so long as were consistent in how we treat the problem. And it's the same here we can divine define zero potential over here on the negative sheet over here on the positive she or at any point in between. And as we work in future problems, we will define potential to be zero at wherever is most convenient for us, and we'll see that in future videos. So next let's consider the potential energy associated with a point charge. We'll do so in the same method will consider how much work it takes to move a test charge around this point charge. Now we know the electric field, and hence the electric force is directed outwards from a point charge. Its magnitude will depend on the radius in the relation. Say we have test charge of little Q over R squared, and this will be directed radio the outward, and I won't draw in all of the field lines, but just enough so we can start working this example. Now let's put our point. Charge little que here, and it's going to take some path up to here so we'll call this point a and this point. B. Well, what's the work done? Work. It's simply the integral from A to B of f dot d l. Now what is this value? Well, let's zoom in on, say, this section here. What does that look like? Well, we know our forces directed this way, and our d l is roughly directed something like that. Well, the dot product is merely the projection of D. L on two F and you'll notice we're moving in the radial direction straight into or in some cases, depending where we are out of the field line. So FDR d l is merely d r. And this makes sense. The force is directed radial e outward from our point charge and so on. Lee, the component of movement in the radial direction should matter for work. And so now we can integrate this as FDR. And of course, we have our expression for are here, which will give us negative Q Okay, Capital Q, little Q over our or negative? Que que que? It was one over B minus one over a. Now, as I mentioned, we can define zero wherever we want it to be for our potential energy. And as we know, work is the negative change in potential energy, which is negative U F plus you initial and is equal to negative K Capital Q Q Overbey plus K capital Que que over a. And so now we see another expression just kind of pop out at us that for a point charge you will be are dependent and his equal do Que que que over our and you'll notice. Even though we didn't do so intentionally, we defined or zero. As our gets bigger and bigger, this term goes to zero. In other words, as we get farther away toward infinity, our potential drops to zero. And so that's how we've defined zero potential. Now. Does this make sense? Well, as we get closer and closer to the denominator gets smaller and our potential energy gets much bigger, and we expect the potential to be much larger, near positive charge. Now say we have some Q minus we swift flip the sign there, we could go through all the work and what will arrive at is our potential energy becomes this. So at infinity we still get zero. But as we get closer and closer, we get a large negative number, meaning we get smaller potential or less potential. So as we expect potential energy, it's smaller near a negative charge than it is farther away from it. So we've shown this is a great expression for the potential energy of a point charge and as well work in future videos will show more examples, and generally we will define zero potential as being at infinity. It just works out very well that way from any many instances, but there'll be unexceptional to that will work out along the way.
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