Automobile drives into a brick wall

  • Thread starter Thread starter bmoore509
  • Start date Start date
  • Tags Tags
    Automobile Wall
Click For Summary
SUMMARY

The discussion centers on calculating the initial speed of a 3000 kg automobile that crashes into a brick wall, utilizing the principles of Hooke's law and energy conservation. The effective spring constant of the bumper is 6 × 106 N/m, and the bumper compresses 4.23 cm during the collision. By equating the work done on the bumper to the kinetic energy of the car, the initial speed is determined to be 1.89 m/s. The solution involves converting the compression distance to meters and applying the formulas for spring potential energy and kinetic energy.

PREREQUISITES
  • Understanding of Hooke's law and spring constants
  • Knowledge of kinetic energy and potential energy equations
  • Ability to convert units (e.g., cm to m)
  • Familiarity with energy conservation principles
NEXT STEPS
  • Study the derivation of Hooke's law and its applications in real-world scenarios
  • Learn about energy conservation in elastic collisions
  • Explore advanced topics in mechanics, such as impulse and momentum
  • Investigate the effects of different materials on energy absorption in collisions
USEFUL FOR

Physics students, automotive engineers, safety test analysts, and anyone interested in the mechanics of collisions and energy transfer in automotive safety tests.

bmoore509
Messages
66
Reaction score
0

Homework Statement



An automobile of mass 3000 kg is driven into a brick wall in a safety test. The bumper behaves like a Hooke’s-law spring. It has an effective spring constant of 6 × 106 N/m, and is observed to compress a distance of 4.23 cm as the car is brought to rest. What was the initial speed of the automobile?
Answer in units of m/s.


Homework Equations


F=-kx
Ke=(1/2)mv^2
U=(1/2)kx^2
Kf+Uf=Ki+Ui
W=Kf-Ki+Uf-Ui

The Attempt at a Solution


I'm really not sure where to start. I tried doing KE=U but I know that's not right.
 
Physics news on Phys.org
The kinetic energy of the car is absorbed by the bumper.
Use the spring formula (F=kx) and spring constant given to calculate the energy "stored" in the bumper when compressed 4.23cm
Energy stored will be average force times distance compressed.
Equate this to the kinetic energy of the car before collision.
 
I got it. I had a way to do it but the answer never seemed right. I didn't think about the fact that I needed to convert the 4.23 cm.

I did:
F=-kx = -253800
W=Fd = -10735.74
W=-.5mv^2+.5kx^2
-10735.74=-.5(3000)v^2-.5(6X10^6)(0.0423^2)
V=1.89m/s
 
Good.
It was just a case of putting ½Fx² equal to ½mv²
Kinetic energy lost by the car is "absorbed" by the bumper.
 

Similar threads

  • · Replies 1 ·
Replies
1
Views
2K
  • · Replies 12 ·
Replies
12
Views
1K
  • · Replies 5 ·
Replies
5
Views
2K
  • · Replies 1 ·
Replies
1
Views
2K
  • · Replies 2 ·
Replies
2
Views
10K
  • · Replies 3 ·
Replies
3
Views
6K
  • · Replies 3 ·
Replies
3
Views
3K
  • · Replies 2 ·
Replies
2
Views
1K
Replies
7
Views
5K
  • · Replies 10 ·
Replies
10
Views
9K