Work and Energy Legacy Problem #30 Guided Solution

A good problem solver reads the problem carefully and develops a mental picture of what's going on, maybe even constructing a diagram of the situation, identifies the known information, maybe even recording it on the diagram, and the unknown quantity. Then, a good problem solver is going to use physics understanding and math relationships in order to plot out a strategy to get from the known to the unknown information. Here in this problem, we're told of a child who's put in a swing and pulled back at an angle of 26 degrees with respect to the vertical, is released, let go of, and then swings to the lowest position and likely past it. What we wish to know is what is the speed of this child at the lowest position in the trajectory. You'll notice a diagram has been provided here, and thankfully so. We notice that the dot represents the child, and there's a cable represented by a diagonal line of length L. That length is 1.8 meters long. The child is let go of and then swings until the cable is at the most vertical orientation, and that's the lowest point in the trajectory. We wish to find the v of the child at that location. Now, we have a five-term equation listed on this page, and in that five-term equation, we can cancel a few of the terms. First, on the left side, we can cancel the W and C term, the work done by non-conservative forces. Tension is certainly a non-conservative force, but it acts perpendicular to the motion of the swinger or the child in all locations along its trajectory, and as such, cannot do any work upon the child. There's going to be assumed negligible air resistance and friction is stated in the problem, so W and C simply ends up being zero. Now, we can also eliminate the KE initial term, because initially the child is released from rest, and so KEI is zero. And so all that we have on the left side of this five-term equation is PEI, or we could state that as MGHI. Now, on the right side of the equation, we have only one term. We're going to presume that this lowest point in the trajectory is the height of zero, and as such, there's no PE, there's no PE final, but only KE final. So our equation has changed into PEI equals KEF, where the PEI is the mass of 14 times the g of 9.8 times the initial height, which we don't know. But we'll be able to find out, because as you see on the diagram, the initial height is that little segment of the vertical line that goes from the bottom-most point in the trajectory to that location where the horizontal line intersects the vertical. We can find that height because from the pivot point of the swing down to that location of the child, that distance is L. And it's divided up into two segments, one which is HI, and the longer one of which is L times the cosine of theta. L cosine theta, because that's the side adjacent to theta on a right triangle that has L as its hypotenuse. Now we can find this L cosine theta by plugging in the 26 degrees into our calculator and the 1.8 meters for the length of the chain. We get 1.6178 meters as L cosine theta. Subtracting that from L gives us the initial height. It comes out to be 0.1822 meters. Now that we know the initial height, we can find the initial potential energy, SMG times this HI, where the M is 14 and the g is 9.8. That gives us 24.9938 joules of potential energy initially, and that's equal to the kinetic energy, finally, according to our reduced form of the equation. Now I set this 24.9938 joules equal to 1 half times 14 times the final velocity squared, and I solve for Vf. When I do, I get 1.8896, and I'll round that to two significant digits, such that it's 1.9 meters per second.