Force to accelerate a RipStik

Using suggestions to my future self from last fall, I turned the run around and moved the run onto a flat surface. I preselected an 800 cm flat stretch in front of the south faculty building.

The distance was 8 meters, the time was 4.88 seconds, the force was 20 Newtons, and the mass for the board and I was 68.7 kg. Note that Wendolin indicated that the force was 20 N, but that is also the maximum for that particular spring scale. There exists the possibility that the scale was maxed out and thus that force could have been above 20.

Using d=½at² one can derive a = 2d/t². Using F = ma, one can derive a=F/m. In theory these shoudl be equal. The reality was that the first equation produced 0.67 m/s² while the second produced 0.29 m/s². The energy demonstration a week earlier suggested losses of up to 50% to friction, so maybe friction is an issue in the above inequality. As noted, I also have questions on the constancy of pull and possible maxing out of the scale. Were I to do this again I would likely do a couple more runs and shoot for something less than scale max. Maybe the key is to try the 30 N scale and try to get the tow to pull a constant 20 N force.

Wendolin was towing, while holding the pad and lugging a laptop.

Equipment needs can be seen scattered on the ground - RipStik, rope, tape measure. Also a stop watch and chalk.

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