Finding Spring Constant/Compression

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In summary: I should probably do it with algebra.In summary, 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. To ensure passenger safety, the acceleration of the vehicle must not exceed 5.00g. The required spring constant k is found by equating the work done by the spring to the kinetic energy of the car. Using this equation, the student calculated a value of 20731 N/m for k, however, the online homework program rejected this answer. To find the distance the spring will compress in slowing the vehicle to a stop, the student used the equation F=ma=k
  • #1
BobBarker444
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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.
 
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  • #2
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).
 
  • #3
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:
  • #4
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, [itex] ma = kx [/itex] and [itex] \frac{1}{2} kx^2 = \frac{1}{2}mv^2 [/itex]. And you have two unknowns, [itex] k [/itex] and [itex] x [/itex]. That's enough to solve the problem with algebra.

Note that the value of the acceleration, [itex] a [/itex] 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 [itex] a [/itex] as a variable: we ask ourselves, "what is the [itex] x [/itex] that causes [itex] a = 5g [/itex], when using the [itex] ma = kx [/itex] equation?" So [itex] a [/itex] is not an unknown. The only unknowns left are [itex] x [/itex] and [itex] k [/itex].
 
Last edited:
  • #5
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, [itex] ma = kx [/itex] and [itex] \frac{1}{2} kx^2 = \frac{1}{2}mv^2 [/itex]. And you have two unknowns, [itex] k [/itex] and [itex] x [/itex]. That's enough to solve the problem with algebra.

Note that the value of the acceleration, [itex] a [/itex] 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 [itex] a [/itex] as a variable: we ask ourselves, "what is the [itex] x [/itex] that causes [itex] a = 5g [/itex], when using the [itex] ma = kx [/itex] equation?" So [itex] a [/itex] is not an unknown. The only unknowns left are [itex] x [/itex] and [itex] k [/itex].

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

1. How can I find the spring constant?

The spring constant can be found by dividing the force applied to the spring by the resulting displacement or compression of the spring. This can be represented by the equation k = F/x, where k is the spring constant, F is the applied force, and x is the displacement or compression.

2. What are the units of spring constant?

The units of spring constant depend on the units used for force and displacement. In the SI system, the units for spring constant are N/m (newtons per meter), while in the English system, the units are lbf/in (pounds-force per inch).

3. How do I measure the compression of a spring?

The compression of a spring can be measured by using a ruler or measuring tape to determine the difference in length between the spring in its relaxed state and when a force is applied to it. The difference in length is equal to the compression of the spring.

4. Can the spring constant change?

Yes, the spring constant can change depending on factors such as the material and dimensions of the spring, as well as temperature and external forces. However, for most practical purposes, the spring constant of a particular spring is considered constant.

5. How can I use the spring constant to predict the behavior of a spring?

The spring constant can be used to predict the behavior of a spring by applying Hooke's Law, which states that the force applied to a spring is directly proportional to the spring's displacement or compression. By knowing the spring constant, we can calculate the force required to compress or stretch the spring to a certain length, as well as the resulting displacement at a given force.

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