Physics NUTS needed!!! PE Elastic Energy...

[MedicineNutt]

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An 86.0 kg climber is scaling the vertical wall of a mountain. His safety rope is made of nylon that, when stretched, behaves like a spring with a spring constant of 1.20 x 10^3 N/m. He accidentally slips and falls freely for 0.750 m before the rope runs out of slack. How much is the rope streched when it breaks his fall and momentarily brings him to rest?

My Approach:
1/2(86)(3.38 m/s) + (86)(9.8)(0) = 1/2(86)(0) + 1/2(1200)x --- i solved from the middle and the bottom

145.3 = 600x

x = .242 m

I found PE and KE at the middle and PE(elastic) and PE at bottom...the answer is wrong for sure though--it doesn't even come close to the book answer at 1.95m

HELPPPPPPPPP!!!! physics sucks...strugglin all year

 

[deleted106503]

The distance he falls is 0.750m + the slack since he's not falling on a concrete slab (unless I understood this part wrong). So if you saw it from the side, he falls down for 0.75 meters, hits the net which is flat, and continues to fall past that until he actually stops (no more slack). I'm using "x" as the slack distance.

Potential energy is (mass)x(gravity)x(height) = (86kg)x(9.8m/s^2)x(0.750m + x)

Set that equal to elastic potential energy = (1/2)(k)(x^2)

Expand: (86)(9.8)(.75) + (86)(9.8)(x) = (1/2)(1.2 x 10^3)(x^2)

632.1 + 842.8x = 600x^2

I solved for that by graphing and finding the x-intercept. 1.946 is X (plug it into my formula, it works). Not sure if there's another way to solve it, but that's how I did it. Hopefully you get to use a graphing calculator, lol!

 

[CareerBaffled]

You can use the quadratic formula to solve for the equation 623.1 + 842.8x = 600x^2. Arrange the equation like:

600x^2-842.8x-623.1=0, then plug into (-b(+/-) sqrt(b^2-4ac))/2a

Obviously a graphing calculator would be a better option if allowed, with ugly numbers like these.

WilliamsF1's solution looks good to me. The potential energy (from gravity) at the top should equal the potential energy (from the "spring") at the bottom. (Keep in mind that the climber is not moving at either of these points, so kinetic energy is zero).