Work and Energy Legacy Problem #20 Guided Solution

Here we have a picture of a pitcher of cola being pushed across a level countertop by an applied force. We would say that work is being done by pizzeria upon the pitcher of cola. Now what we're asked to do ultimately is to determine the speed of the cola when it finally is done being pushed upon. In order to do this, what we need to do is calculate the work done by P and the work done by friction, and thus the total work. And equate this total work with a kinetic energy change. That will be our basic overview of the solution. Fortunately for us, the problem steps us through in a four-step process in order to get to the point of actually calculating the speed of the pitcher. The first step is to determine the work done by P upon this pitcher as he pushes at a distance of 48 centimeters. We need to use the equation work is equal to Fd cosine of theta, where the theta is the angle between F and D. In this case, for Pete's applied force, it happens to be 0 degrees. So we need to take the 8.8 newtons and the 48 centimeters and use these figures for F and D in order to calculate the work. The first complication has to do with this D value. It's in centimeters, so we need to change it to .48 meters. And we need to go 8.8 times .48 times the cosine of 0 degrees, and we get 4.224 joules of work being done. In part B, we're asked to determine the work done by the friction force upon the mug. We're told that the coefficient of friction is .28. We need to find from that the force of friction. So if we think in terms of a free body diagram representing the forces acting upon this pitcher, what we would have is a down force of gravity and an up normal force from the countertop pushing up. These would both be calculated as m, the mass, times g, 9.8 newtons per kilograms. And the friction force is always mu F norm. And so we would be going mu times F norm to get the friction force, or mu times mg. Substituting in for m, 2.6 kilograms, and 9.8 for g, and .28 for the coefficient of friction gives us a friction force of 7.1344 newtons. Now we can calculate the work done by this friction force. We just go F frict times d times the cosine of the angle between F frict and d, which here would be 180 degrees. So we go 7.1344 newtons multiplied by .48 meters, and then multiplied by the cosine of 180, and that gives us a negative 3.4245 joules of work. Now we've determined the work done by P, which is a positive value, the work done by friction, which is a negative value, and now to determine the total work in part C, we have to add these two quantities together. Adding the negative to the positive gives us 0.799 joules of work being done. We'll round this to the first decimal place. That's .8 joules of work being done, and it's positive. This .8 joules of work acts upon the picture of cola in order to accelerate it from rest to some final speed, or to some final kinetic energy, and we could say that the work done is equal to the kinetic energy change. The kinetic energy change is .8 joules, or .799 joules, and so what we have is an initial kinetic energy of zero, and a final kinetic energy of .799 joules, and that's the answer to D. We need to use this answer to D in order to calculate the speed of the pitcher when Pete's finally done pushing up on it. So we equate the .799 joules to one-half mv squared, one-half times 2.6 times v squared. Solving that for v, we get .784 meters per second. We'll round that to the first decimal place, it becomes .8 meters per second.