also b. Calculate the unstretched length of the bungee cord.
A bungee jumper with mass 64.0 kg jumps from a high bridge. After reaching his lowest point, he oscillates up and down, hitting a low point eight more times in 34.5 s. He finally comes to rest 24.0 m below the level of the bridge.
2006-12-07 12:40:34 · 2 answers · asked by Layla 1 in Science & Mathematics ➔ Engineering
The frequency of the oscillation is dependent on the mass and the spring rate. I think the equation is something like period equals the square root of spring rate over mass (look it up)
You know the period, and you know the mass. Solve for the rate. Then use the rate and the weight of the jumper to determine the "stretch" in the cord. Subtract this from the 24m and you will have the "unstretched" length.
2006-12-07 12:59:17 · answer #1 · answered by www.HaysEngineering.com 4 · 0⤊ 0⤋
not sufficient ideas... replaced into the bungee twine totally prolonged at 40 m or Did the jumper circulate SPLAT on the floor 40 m decrease than? If the jumper went splat (or in any different case touched the floor), you could not practice Hooke's regulation because of the fact the spring did not totally counteract g. i.e.; you could not calculate the spring consistent. F = -kx
2016-12-11 04:30:31 · answer #2 · answered by Anonymous · 0⤊ 0⤋