How much work does tension do to pull the mass from the bottom of the hill (θ = 0) to the top at constant speed? To answer this question, write an expression for the work done when the mass moves through a very small distance ds while it has angle θ, replace ds with an equivalent expression involving R and dθ , then integrate.

Video transcript

Hey, everyone. So this problem is about work and forces. Let's see what it's asking us. A flat section of the earth's surface has a depression with straight and parabolic walls. Determine the work done by a force that pulls a mass from the lowest point in the depression where theta equals zero to a point out of the depression at a constant speed. Then they give us a hint, work out an expression for work done in moving the mass through a tiny distance given by D L when its orientation is theta, then substitute for D L with an appropriate expression with R and delta theta. Before integrating our multiple choice answers here are a M G R B negative M G R C M G or D M G D L. OK. So the first thing we're going to do here is draw our free body diagram. So we have this mass, we have the force working on the mass to pull it out of the depression. So that's gonna be in the positive Y direction. And then we have weight acting in the negative y direction from Newton's second law. We can recall the sum of the forces is equal to the mass multiplied by the acceleration. Now, in the problem, we're told that the force is pulling at a constant speed Which tells us that the acceleration is zero. So our force minus our weight is equal to zero, we can rewrite that as force is equal to the weight, the weight we can recall is given by mass multiplied by gravity multiplied by the cosine of data. The next step in this problem because they're asking for work is to recall the relationship between work and force. So work is equal to force multiplied by the distance. So the differential of work done when mass is pulled through an angle delta, theta corresponds with the distance R delta theta. So we can write that as T equals R delta theta. And so we can then rewrite this work equation in the term of the integral Work equals the inch curl from zero to pie over two radians of mass gravity cosine theta multiplied by R D theta. So when we take this integral, we are left with work equals M G R sign data from 02 Pi over two radiant. And so we'll solve that out and that will be M G R multiplied by a sign of pi over two minus M G R multiplied by sign of zero. So the sign of zero we can recall is zero. So that term goes to zero. And then the sign of pi over two radiations is one. So that simplifies to work equals M G R. So that is the answer to this problem and that aligns with answer choice A so that's all we have for this one. We'll see you in the next video.