two.twenty-five.09

Today Mr. Manning felt as thought to he needed to get in touch with his inner child (ha). With this feeling he had a student hepl in carrying his ginormous K'NEX roller coaster. Giving todays lession a little twist, it was mostly an open class discussion. He first started off by setting up the gates that the cart would be traveling through making sure that there were no problems that could occur during the procedure. When everything was finally in its right position we bagan to collect our data. We started by finding the total amount at the top of the coaster by using the equation Ep(i)= mgh. Next we then found Ep(f) and after that was solved we solved the equation Ek(f)= 1/2mv^2 in order to come up with the coasters kinetic energy. Here is where we ran into a problem though, what was the velocity of the moving object? This is where the two gates, that Manning set up, came into play. By finding our delta x (.273 m) and our time (.1242 s) we were able to discover that our v(f)= 2.20 J. Once the velocity was found we then took the answer and pluged it into the equation. Work shown for further understanding: Ep(i)= mgh =.096 kg X 9.8 m/s^2 X .957 m =.900 J

Ep(f)=mgh =.096 kg X 9.8 m/s^2 X .126 m =.119 J

Ep(f)= 1/2mv^2 { delta x(.273 m) divided by time (.1242 s)= 2.20 J } =1/2 X .096 X (2.20 J) ^2 =.232 J

Next we needed to find the amount of percent lost. To do so we used the equation % lost = Etot(i) - Etot (f) X 100 divided by the total amount of energy. = .900 J - .351 J X 100 = 54.9 / .900 = 61 %

When all of this was completed Mr. Manning instructed the class to finish up on the Energy and Roller Coasters paper (page31) and the K'NEX Roller Coaster Demo (page 35), which was given to the class in the beginning of the period.