x = 1/2*g*t^2 = 2 = 4.9*t^2
Solve for t, the time then solve for the climber's speed v using
v = 9.8*t
1/2*m*v^2 = 1/2*k*xstretch^2; where xstretch is the amount that the rope will stretch
How much would the rope stretch to break the climber's fall if he free falls 2m before the rope runs out of slack? k=1.40x10^4N//m
mass of man= 90 kg
I forgot how to use conservation of energy for this. I tried it, but I did it wrong.
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