Work and Energy Legacy Problem #25 Guided Solution

An effective problem solver reads the problem carefully and develops a mental picture of what's going on, records the known information and the unknown quantity, and then uses physics ideas and principles in order to plot out a strategy for getting from the known to the unknown quantity. This problem describes a lab environment in which there's a cart up on a ramp. The cart is given an initial push, accumulating a speed of 2.35 meters per second at a height of 0.125 meters above the lab table. The cart rolls up the incline and it comes to a stop, but at the height of 0.340 meters, which is not necessarily the stopping position, we're asked to determine the speed of the cart. You'll notice there's a five-term equation here on this page, and we're going to use that equation in order to try to determine the final speed of the cart. Using the equation involves canceling terms which are negligible. Initially, we do have kinetic energy. It's based upon the speed of 2.35 meters per second, and initially, there's some height, at least height relative to the lab table. When it gets to a height of 0.340 meters above the lab table, it clearly has potential energy, and likely, it has kinetic energy as well, since they're asking us to calculate the speed. The one term in the equation that cancels is the term WNC. This term cancels because the only force doing work is the force of gravity, and it's a conservative force. The normal force acts perpendicular to the motion at a 90-degree angle, and the FD cosine of the theta for this force is zero. So to solve for this final speed, I'm going to begin to calculate individual terms within this equation. The first term, MGHI, can be calculated if we take the mass of the cart, 1.0 kilograms, multiply by 9.8, and then multiply by the height, 0.125 meters. It gives me a value of 1.2250 joules. I'm going to write that down. The kinetic energy term can be canceled if we go 1 half times the mass of the cart, 1 kilograms, multiply by the initial speed at 2.35 meters per second. Doing so gives me 2.7613 joules of energy. Initially, the total amount of mechanical energy can be found by summing these two terms. It comes to 3.9863 joules of energy. When the cart gets to this height, the 0.340 meters above the lab table, it will still possess 3.9863 joules of energy. Some of it will be in the form of potential, and some in the form of kinetic. If we wish to calculate the speed of the object, we need to find out the kinetic energy. Doing so demands that we take this 3.9863 and set it equal to MGHF plus this kinetic energy finally. The MGHF at the height of 0.340 meters can be calculated by going 1 half times the mass of 1 times this height of 0.340 meters. It comes out to be 3.332 joules of energy. We can subtract 3.332 joules of energy from the initial amount, and that would give us the total amount of kinetic energy at this particular height. That comes out to be 0.6543 joules. We can set 0.6543 joules equal to 1 half MVF squared, where the M is 1 kilogram, and then we can solve for VF. Doing so gives us 1.1439 meters per second, and we'll round that to two significant digits.