A mountain climber of mass 60 kg slips and falls a distance of4m, at which time

Question

A mountain climber of mass 60 kg slips and falls a distance of4m, at which time he reaches the end of his elastic safety rope.the rope then stretches an additional 2m before the climber comesto rest. What is the spring constant of the rope, assuming it obeysHooke's law? My teacher gave me the answer as 1760 N/m, but I don'tunderstand how to get that. I set up 2k=(60)(9.8) but feel (and know) that I leftsomething out. A mountain climber of mass 60 kg slips and falls a distance of4m, at which time he reaches the end of his elastic safety rope.the rope then stretches an additional 2m before the climber comesto rest. What is the spring constant of the rope, assuming it obeysHooke's law? My teacher gave me the answer as 1760 N/m, but I don'tunderstand how to get that. I set up 2k=(60)(9.8) but feel (and know) that I leftsomething out.

Explanation / Answer

We know that    Workdone = U = 0.5kx2 =mg* (d+x) ==> k = mg (d+x) / 0.5x2            = 6 0*9.8 * 6 / 0.5 * 22            = 1764N/m

Related Questions

Navigate

• Browse (All)
• Browse A
• Subjects

---