For a given spring of length L and spring constan k, how does k change if I cut the spring in half?

Mountain climbers use nylon safety rope whose elasticity plays an important role in cushioning the sharp jerk if a climber falls...



(a) Suppose that a climber of 80 kg attached to a 10-m rope falls freely from a height of 10 m above the point at which the rope is anchored to a vertical wall of rock, to a height 10 m below the point at which the rope is anchored. Treating the rope as a spring with k=4.9*10^3 N/m, calculate the maximum force that the rope exerts on the climber during stopping.



(b) Repeat the calculations for a rope of 5.0 m and an initial height of 5.0 m. Assume the rope is made out of the same material as the first, and remember to take into account the change in the spring constant due to the change in length. compare your results for (a) and (b).For a given spring of length L and spring constan k, how does k change if I cut the spring in half?
I agree with 'railrule' for the first part only.

a)

The energy stored in the spring = the change in p.e.

k1*x^2 / 2 = mgH

k1^2* x^2 = 2k1*mgH

Force exerted by the rope F = kx

F1 = k1*x = 鈭? 2k1*mgH) = 鈭?2*4.9*10^3*80*9.8*20) = 1.24*10^4 N



b)



k = 蟺nr^4l / 2LR^2

Spring constant k is inversely proportional to length of the spring L



F2 = k2*x = 鈭? 2k2*mgH' )

H' = 10 m

k2 = 2k1

F2 = 鈭?2*2*4.9*10^3*80*9.8*10) = 1.24*10^4 N



F2 = F1For a given spring of length L and spring constan k, how does k change if I cut the spring in half?
hi



Energy equation



Spring energy = potential energy

(1/2) k x^2 = m*g*H



k x^2 = 2*m*g*H

k^2 x^2 = 2*k*m*g*H



Fsp = kx = sqrt (2*k*m*g*H)



Fsp = sqrt ( 2* (4.9*10^3 N/m) * 80 kg * (9.8 m/s^2) * 20 m)

Fsp =1.24 10^4 N



b)

Fsp = sqrt ( 2* (4.9*10^3 N/m) * 80 kg * (9.8 m/s^2) * 10 m)

Fsp = 8.77 10^3 N



Proportion is 1.41 = sqrt(2)



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