CALC The potential energy for a particle that can move along the x -axis is U=Ax²+B sin(πx/L) , where A , B , and L are constants. What is the force on the particle at (a) x=0 , (b) x=L/2 , and (c) x=L?

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Hey, everyone in this problem, we're asked to imagine a ball rolling along a horizontal track under the influence of potential energy given by U of X is equal to A X squared plus B cosine of pi divided by two, L multiplied by X where A B and L are constant. We're asked to determine the magnitude and direction of the force acting on the ball. When it reaches part one, the starting point X equals zero, part two, the midpoint of the track, X equals L divided by two and part three. The end of the track at X equals L. We're given four answer choices A through D each containing a different value for the force F for the three time points that we are interested in. We're gonna come back to these answer choices as we work through the problem. Now, let's get started. We're being asked for a force. We're given information about the potential energy. So let's recall the relationship between those two things. The force F going to be equal to negative the derivative of the potential energy with respect to X. So negative DU divided by DX and we can just write this as negative D by DX of our potential energy function, which is A X squared must be cosine of pi divided by two L multiplied by X. All right. So let's go ahead, we're gonna take the derivative of this and figure out what our force is gonna be. And once we get an equation for the force, we can substitute in our X values of interest at those three points and solve for the force. So taking the derivative with respect to X of our potential energy function, the first term we have is A X squared. OK. So the X squared taking the derivative, this is an exponent. We're gonna bring that exponent down in front and then reduce the exponent by one. OK. Now don't forget this negative up front. We're gonna leave it out in front and put big square brackets. So we remember to apply it to everything. OK. So we have negative two outside of the brackets for that negative. There we go. So we have negative and then in brackets, we have two a multiplied by X. The derivative of cosine is going to be negative sine that B is just a constant. So it's gonna stay out in front. So we get negative B multiplied by sine, everything that was inside of our cosine function. So pi divided by two L multiplied by X and then by the chain rule, we have to multiply by the derivative of what's inside of that cosine function. And inside we have pi divided by two L multiplied by X. OK. So a constant multiplied by X, we take the derivative of that and we just get that constant. OK. So just the pi divided by two L good. Now let's go through and simplify, we have that our force F is going to be negative two multiplied by A X plus pi divided by two L multiplied by B multiplied by sine of pi divided by two LX. OK. So we've just simplified rearrange some of those constants so that they are together. And now we can get to this question answering each part. OK? So for part one, we wanna look when X is equal to zero, we're gonna take that force function that we just found, we're gonna substitute X is equal to zero. So we're gonna look at F of zero, we get negative two multiplied by a multiplied by zero plus pi divided by two L multiplied by B multiplied by sine of zero. A sign of zero is just zero. Anything multiplied by 00. We just get zero plus zero and we get a force of zero. But let's take a look at our answer choices. Options A and B have a nonzero force at 0.0 X equals zero. OK. So we can actually eliminate those two answer choices. Option C and D both have that force being zero, which is what we want. So that's good. Now we're gonna move to the next age and we are going to do the midpoint of the chart. OK. So when X is equal to L over two, and again, we're gonna substitute that X value into the force function we found. So F L divided by two, it's going to be equal to negative to a multiplied by L divided by two plus pi divided by two. L multiplied by B multiplied by sine of pi divided by two L multiplied by L divided by two. All right, let's go ahead and simplify. So in our first term, these two are going to divide it and we're gonna be left with negative A L. In the second term, we have our pi divided by two L multiplied by B. We can't do anything with that yet. We're multiplying by sine and inside of the bracket. These LS are going to divide it and we're gonna be left with sine of pi divided by four. Now sine of pi divided by four. Recall is equal to one divided by the square root of two. We can rationalize that denominator multiply the numerator and denominator by the square root of two. And we can write that as the square root of two divided by two. OK. So let's go ahead and write it in that way. And we get that the force at the midpoint is gonna be negative A L plus pi divided by two L multiplied by B multiplied by the square root of two divided by two. All right. So we're gonna leave it like that for now and we are going to now move on to part three. OK? And part three is at the end of the track where X is equal to L. So let's set this up again. We're using that same for force function and we are substituting X is equal to L wait. Yeah, negative two A L plus Pi divided by two L multiplied by B multiplied by sine of pi divided by two L multiplied by L A factor of L inside the brackets is going to divide. Oh, so we're gonna be left with negative two A L plus Pi divided by two L multiplied by B multiplied by sign of pi divided by two. Can I recall that Pi divided by two is just equal to one? And so we can simplify this one step further and write this as negative two multiplied by A L plus pi divided by two L multiplied by B. And so now we have our four forces or sorry, our three forces at the beginning, midpoint and end. OK. So at the beginning, the force is zero at the midpoint, negative A L plus pi divided by two L multiplied by B multiplied by the square root of two divided by two. And at the end negative two A L plus Pi divided by two L multiplied by B. Let's take a look at our answer choices, we narrowed it down to C or D after we looked at the starting point. Ok. So looking at part two, we can see that option C has the correct force function for part two. And it also has the correct answer for part three. Ok. Option D did have the correct answer for part three, but it did not have the correct answer for part two. Ok. And so the correct answer here is c thanks everyone for watching. I hope this video helped see you in the next one.

CALC The potential energy for a particle that can move along the x -axis is U=Ax²+B sin(πx/L) , where A , B , and L are constants. What is the force on the particle at (a) x=0 , (b) x=L/2 , and (c) x=L?

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Key Concepts

Potential Energy

Force and Potential Energy Relationship

Differentiation