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Work-Energy Theorem Question

  • Thread starter kylepetten
  • Start date
  • #1
25
0

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] [Broken] = ?

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] [Broken]

= [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:

Answers and Replies

  • #2
rock.freak667
Homework Helper
6,230
31
65 km/h ≠ 234 m/s

1h = 3600 s.

Convert it over.
 
  • #3
25
0
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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