By Laura Aseltine
Though I usually take the cruiser up the hill for class in the morning, lately I’ve been brainstorming some better ways to ride in style. One option I’ve been considering is to have my kind classmate Melissa Barnard pull me up the hill in a wagon. It can’t be that much work can it? I decided to use my nifty new physics calculate.
To calculate the work I decided to split it up into Whorizontal and Wvertical. Since W=F//d I first needed to consider the forces in the horizontal direction, which would be Fmelissa and Ffr. (I’m assuming that the only nonconservative force is Ffr). I decided that we would be accelerating at 0.05 m/s2 and Melissa would be pulling me at a 15° angle. The force that she is pulling at is F=ma so I would need the mass. I added the mass of the wagon (13 kg), the mass of me (80kg) and the mass of my backpack (10kg) to get a total mass of 103 kg. F= 103kg * 0.05 m/s2 = 5.15 N. To determine the horizontal component of Fmelissa I calculated cos(15°)*5.15= 4.97 N. I decided that the force of friction would be 0.5 N.
So to calculate the Whorizontal I took the sum of the forces in the parallel direction (4.92N-0.5N) and multiplied it by the displacement. I used mapmyrun.com to map the displacement from my townhouse to the Ho. It was 0.8 miles, which converts to 1300 m. Therefore Whorizontal = (4.97N-0.5N)*1300= 5811J.
I then calculated the Wvertical. The only force in the vertical direction is the y component of FMelissa. I calculated this to be sin(15°)*5.14=1.33N. I multiplied this by the displacement in the y direction. Again I used mapmyrun.com to get the change in elevation, which was 197 ft, which converts to 60. m. Therefore Wvertical = 1.33N*60.m=79.8J.
To get the Wnet, I added these two works together. Wnet = Whorizontal + Wvertical = 5811J + 79.8J = 5889.9 J. Maybe this will convince her! Anyone have a wagon?