Finding Spring Constant/Compression

  • Thread starter Thread starter BobBarker444
  • Start date Start date
  • Tags Tags
    Spring
Click For Summary

Homework Help Overview

The discussion revolves around a physics problem involving a spring and a vehicle's deceleration during a crash scenario. The original poster attempts to find the spring constant and the distance the spring compresses when a vehicle comes to a stop after colliding with the spring. The problem is situated within the context of dynamics and energy conservation.

Discussion Character

  • Mixed

Approaches and Questions Raised

  • Participants discuss the use of kinematic equations, questioning their applicability due to the non-constant acceleration involved in the scenario. Some suggest using force equations instead, while others explore the relationship between force, mass, and spring compression.

Discussion Status

There is an ongoing exploration of the problem with various approaches being discussed. Some participants have provided hints and alternative equations to consider, indicating a productive dialogue without a clear consensus on the correct solution yet.

Contextual Notes

Participants note the constraints of the problem, including the requirement that the acceleration must not exceed 5g and the assumption of negligible mass for the spring. There is also mention of the online homework system rejecting certain answers, which adds to the complexity of the discussion.

BobBarker444
Messages
1
Reaction score
0
We have online homework that checks our answers when input. I'm pretty confident about this problem, but it won't take my answers.

1. Homework Statement

A student proposes a design for an automobile crash barrier in which a 1350-kg sport utility vehicle moving at 25.0m/s crashes into a spring of negligible mass that slows it to a stop. So that the passengers are not injured, the acceleration of the vehicle as it slows can be no greater than 5.00g.

a) Find the required spring constant k. In your calculation, disregard any deformation or crumpling of the vehicle and the friction between the vehicle and the ground.

b) Find the distance the spring will compress in slowing the vehicle to a stop.

Homework Equations



W = (1/2)kx^2 (work done by spring)
v^2 = vo^2 + 2a(x) (to find x given a)

The Attempt at a Solution


[/B]
I used 5.00(g) = 5(9.8) = 49 to find the maximum acceleration. I plugged this into v^2 = vi^2 + 2a(x) and got x = 6.38 m. This should be the answer to b), but the online homework program says it is wrong.

I see no way to solve a) without b). I used W=(1/2)kx^2 (work done by a spring), set equal to (1/2)mv^2 (the kinetic energy of the car). Solving for k and plugging in values, I got 20731 N/m. The online homework program also rejected this answer.
 
Physics news on Phys.org
Hello BobBarker444, Welcome to Physics Forums! :)

BobBarker444 said:
We have online homework that checks our answers when input. I'm pretty confident about this problem, but it won't take my answers.

1. Homework Statement

A student proposes a design for an automobile crash barrier in which a 1350-kg sport utility vehicle moving at 25.0m/s crashes into a spring of negligible mass that slows it to a stop. So that the passengers are not injured, the acceleration of the vehicle as it slows can be no greater than 5.00g.

a) Find the required spring constant k. In your calculation, disregard any deformation or crumpling of the vehicle and the friction between the vehicle and the ground.

b) Find the distance the spring will compress in slowing the vehicle to a stop.

Homework Equations



W = (1/2)kx^2 (work done by spring)
v^2 = vo^2 + 2a(x) (to find x given a)

The Attempt at a Solution


[/B]
I used 5.00(g) = 5(9.8) = 49 to find the maximum acceleration. I plugged this into v^2 = vi^2 + 2a(x) and got x = 6.38 m. This should be the answer to b), but the online homework program says it is wrong.

I see no way to solve a) without b). I used W=(1/2)kx^2 (work done by a spring), set equal to (1/2)mv^2 (the kinetic energy of the car). Solving for k and plugging in values, I got 20731 N/m. The online homework program also rejected this answer.

For what it's worth, I came up with the same answers that you did (ignoring any minor rounding differences).
 
You cannot use ##v^2 = v_o^2 + 2ax##, as acceleration is not constant. (Recall derivation of SUVAT equations.)

Equation for part (a) Looks alright.

Equation for part (b): Hint - use ##F = ma = kx##

Code:

Code:
Solve[{1/2*1350 * 25^2 ==  1/2 k x^2 && 5*9.81 ==  k x/1350}, {k, x}]

Using Mathematica, I obtained ##k = 5196.75 N m^{-1}##, ##x = 12.7421 m##.
 
Last edited:
Doh! o:) <slaps self on forehead>

Yes, like unscientific said, the kinematics equations for uniform acceleration do not apply here. 'Looks like we both made that mistake, BobBarker444.

However, you shouldn't need to use Mathematica to solve this. Just realize that you have two equations, ma = kx and \frac{1}{2} kx^2 = \frac{1}{2}mv^2. And you have two unknowns, k and x. That's enough to solve the problem with algebra.

Note that the value of the acceleration, a varies when the car is still moving and the spring has not reached its maximum compression. We know that the force on the car, thus the car's acceleration, increases the more the spring is compressed. But when the spring reaches the compression necessary to cause the acceleration to be 5g, we do not need to treat a as a variable: we ask ourselves, "what is the x that causes a = 5g, when using the ma = kx equation?" So a is not an unknown. The only unknowns left are x and k.
 
Last edited:
collinsmark said:
Doh! o:) <slaps self on forehead>

Yes, like unscientific said, the kinematics equations for uniform acceleration do not apply here. 'Looks like we both made that mistake, BobBarker444.

However, you shouldn't need to use Mathematica to solve this. Just realize that you have two equations, ma = kx and \frac{1}{2} kx^2 = \frac{1}{2}mv^2. And you have two unknowns, k and x. That's enough to solve the problem with algebra.

Note that the value of the acceleration, a varies when the car is still moving and the spring has not reached its maximum compression. We know that the force on the car, thus the car's acceleration, increases the more the spring is compressed. But when the spring reaches the compression necessary to cause the acceleration to be 5g, we do not need to treat a as a variable: we ask ourselves, "what is the x that causes a = 5g, when using the ma = kx equation?" So a is not an unknown. The only unknowns left are x and k.

Haha, I am generally lazy so I used mathematica to solve it.
 

Similar threads

  • · Replies 5 ·
Replies
5
Views
1K
  • · Replies 12 ·
Replies
12
Views
3K
  • · Replies 12 ·
Replies
12
Views
2K
  • · Replies 3 ·
Replies
3
Views
1K
  • · Replies 17 ·
Replies
17
Views
3K
  • · Replies 22 ·
Replies
22
Views
2K
  • · Replies 3 ·
Replies
3
Views
2K
  • · Replies 17 ·
Replies
17
Views
2K
Replies
24
Views
4K
Replies
9
Views
2K