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 W

_{horizontal }and W_{vertical}. Since W=F_{//}d I first needed to consider the forces in the horizontal direction, which would be F_{melissa }and F_{fr.}(I’m assuming that the only nonconservative force is F_{fr})._{ }I decided that we would be accelerating at 0.05 m/s^{2}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/s^{2}= 5.15 N. To determine the horizontal component of F_{melissa }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 W

_{horizontal}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 W_{horizontal }= (4.97N-0.5N)*1300= 5811J.
I then calculated the W

_{vertical.}The only force in the vertical direction is the y component of F_{Melissa}. 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 W_{vertical}= 1.33N*60.m=79.8J.
To get the W

_{net}, I added these two works together. W_{net}= W_{horizontal}+ W_{vertical}= 5811J + 79.8J = 5889.9 J. Maybe this will convince her! Anyone have a wagon?
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