Work-Energy Theorem Question

In summary, a skier with a mass of 85 kg and a speed of 65 km/h (234 m/s) crashes into netting at the bottom of a ski hill. The netting, with a "spring" constant of 13500 N/m, brings the skier to rest and does 2.3 x 10^6 J of work. The netting is stretched 18 m when the skier comes to rest.
  • #1
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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] [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!
 
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  • #2
65 km/h ≠ 234 m/s

1h = 3600 s.

Convert it over.
 
  • #3
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!
 

1. What is the Work-Energy Theorem?

The Work-Energy Theorem is a principle in physics that states that the work done by all forces acting on a body is equal to the change in kinetic energy of the body. In other words, the net work done on an object will result in a change in its speed or direction.

2. How is the Work-Energy Theorem used in real life?

The Work-Energy Theorem has many practical applications, such as calculating the energy required to lift an object, determining the speed of a moving object, or analyzing the efficiency of a machine. It is also used in fields such as engineering, sports, and transportation.

3. Can the Work-Energy Theorem be applied to all types of motion?

Yes, the Work-Energy Theorem can be applied to all types of motion, including linear, rotational, and oscillatory motion. It is a general principle that applies to all objects, regardless of their size, shape, or speed.

4. What are the limitations of the Work-Energy Theorem?

The Work-Energy Theorem assumes that there are no external forces acting on the system, and all the work done is converted into kinetic energy. In reality, there are often other factors, such as friction and air resistance, that can affect the motion of an object and result in a loss of energy.

5. How is the Work-Energy Theorem related to the Law of Conservation of Energy?

The Work-Energy Theorem is closely related to the Law of Conservation of Energy, which states that energy cannot be created or destroyed, only converted from one form to another. The Work-Energy Theorem is a specific application of this law, as it shows how energy can be transformed from work to kinetic energy.

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