What is the Work-Energy Theorem for a Skier Crashing into Netting?

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SUMMARY

The discussion centers on the application of the Work-Energy Theorem to a scenario involving a skier with a mass of 85 kg crashing into netting at the bottom of a ski hill. The skier's initial speed is 65 km/h, which converts to approximately 18.06 m/s, not 234 m/s as initially calculated. The work done by the netting to bring the skier to rest is calculated to be 2.3 x 10^6 J. Additionally, using the spring constant of the netting at 13500 N/m, the netting stretches approximately 18 m when the skier comes to rest.

PREREQUISITES
  • Understanding of the Work-Energy Theorem
  • Knowledge of kinetic energy (KE) and potential energy (PE)
  • Familiarity with spring constants and Hooke's Law
  • Ability to perform unit conversions, specifically from km/h to m/s
NEXT STEPS
  • Study the derivation of the Work-Energy Theorem in physics
  • Learn about kinetic energy calculations and their applications
  • Explore Hooke's Law and its implications in real-world scenarios
  • Practice unit conversion techniques for speed and other measurements
USEFUL FOR

This discussion is beneficial for physics students, educators, and anyone interested in understanding the principles of energy transfer and mechanics in real-world applications, particularly in sports and safety equipment design.

kylepetten
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Homework Statement



A skier comes crashing into the netting at the bottom of a ski hill. The skier has a mass of 85 kg and is moving at a speed of 65 km/h (234 m/s).

(a) How much work is done by the netting while bringing the skier to rest?
(b) If the "spring" constant for the netting is 13500 N/m, how far is the netting stretched when the skier comes to rest?

m = 85 kg
v1 = 234 m/s
v2 = 0 m/s
k = 13500 N/m
w = ?
[PLAIN]http://upload.wikimedia.org/math/f/5/d/f5d889f32d6794e1bc2ed394e9688c76.pngx[/URL] = ?

Homework Equations



(F)(d) = [1/2(m)(v2^2)] - [1/2(m)(v1^2)]
KE = [1/2(m)(v^2)]
PEe = [1/2(k)(x^2)]

The Attempt at a Solution



(a)

W = [PLAIN]http://upload.wikimedia.org/math/f/5/d/f5d889f32d6794e1bc2ed394e9688c76.pngKE[/URL]

= [1/2(m)(v^2)]
= [1/2(85 kg)(234 m/s)^2]
= 2327130
= 2.3 x 10^6 J

This answer seems reasonable.

(b)

Now here is where I need help.

2.3 x 10^6 J = [1/2(k)(x^2)]
2.3 x 10^6 J = [1/2(13500 N/m)(x^2)]
x = sqrt{[-2.3 x 10^6 J]/[1/2(13500 N/m)]}
x = 18 m

Does this look right?
Thanks in advance!
 
Last edited by a moderator:
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65 km/h ≠ 234 m/s

1h = 3600 s.

Convert it over.
 
rock.freak667 said:
65 km/h ≠ 234 m/s

1h = 3600 s.

Convert it over.

Ahh! I multiplied by 3.6 instead of dividing! Thanks!
 

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