Spring constant and maximum acceleration

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SUMMARY

The spring constant (k) required to bring a 1150 kg car to rest from a speed of 99 km/h, ensuring that occupants experience a maximum acceleration of 5.0 g, is calculated to be 7318.1 N/m. The process involves converting the initial velocity to meters per second, determining the negative acceleration as -49 m/s², and calculating the stopping distance using the formula V² = V₀² + 2ax. The final calculation utilizes Hooke's Law, F = -kx, to derive the spring constant.

PREREQUISITES
  • Understanding of Newton's laws of motion
  • Familiarity with Hooke's Law
  • Knowledge of basic kinematic equations
  • Ability to convert units (e.g., km/h to m/s)
NEXT STEPS
  • Study the derivation and application of Hooke's Law in mechanical systems
  • Learn about kinematic equations for uniformly accelerated motion
  • Explore the effects of different spring constants on vehicle safety systems
  • Investigate methods for calculating forces in dynamic systems
USEFUL FOR

Mechanical engineers, automotive safety designers, and physics students interested in vehicle dynamics and spring mechanics will benefit from this discussion.

cgward
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What should be the spring constant k of a spring designed to bring a 1150 kg car to rest from a speed of 99 km/h so that the occupants undergo a maximum acceleration of 5.0 g?


I am unsure of the formula to find the spring constant by determining a maximum acceleration of 5 g
 
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What have you tried so far? Can you think of any formulas that might be relevant?
 
PE= 1/2 k x^2
 
cgward said:
What should be the spring constant k of a spring designed to bring a 1150 kg car to rest from a speed of 99 km/h so that the occupants undergo a maximum acceleration of 5.0 g?


I am unsure of the formula to find the spring constant by determining a maximum acceleration of 5 g

This is how you should do it:

1- change the initial velocity to m/s
2- calculate acceleration: 5*9.8m/s^2 = -49m/s^2: note acceleration is negative to stop the car.
3- calculate the distance (x) over which the car come to halt: V^2= V^2 (initial)+2ax (V= final velocity = 0, x= distance, a=acceleration)
4- now you have 'x', calculate 'k' like this: F=-kx and F=ma (where m=mass)

Solution: m=1150kg, V(initial)=99km/h*(1000m/km)*(h/3600s)= 27.5m/s, V(final)=0, now:

0 = (27.5)^2 + 2*(-49)* x, x= 7.7m

now we can calculate 'k': ma = -kx (this come from F = -kx hook's law), so

1150kg*-49m/s^2 = -k * 7.7m, k = 7318.1 N/m

Note: I did this calculation quickly and I might have errors but the method should be all right.

Good luck!
 

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