# Hookes LawUsing Hooke’s Law, we can show that the work done in compressing a spring a distance of x feet from its at-rest position is `W_(s)=1/2kx^2` , where k is a stiffness constant depending...

Hookes Law

Using Hooke’s Law, we can show that the work done in compressing a spring a distance of x feet from its at-rest position is
`W_(s)=1/2kx^2` , where k is a stiffness constant depending on the spring. It can also be shown that the work done by a body in
motion before it comes to rest is given by `W_m=w/(2g)v^2` where w = weight of the object (lb), g= acceleration due to gravity (32.2 ft/sec2), and v =object’s velocity (in ft/sec). A parking garage has a spring shock absorber at the end of a ramp to stop runaway cars. The spring has a stiffness constant k= 9450 lb/ft and must be able to stop a 4000-lb car traveling at 25 mph. What is the least compression required of the spring? Express your answer using feet to the nearest tenth. [Hint: Solve `W_s>W_m` , x ≥ 0].

txmedteach | High School Teacher | (Level 3) Associate Educator

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At the minimum compression of the spring, the work done by the spring will be equivalent to the work done by the moving vechicle. This allows us to set our equations equal to each other based on conservation of energy:

`W_s = W_m`

`1/2kx^2 = w/2gv^2`

Now, we would be able to add in the values, but keep in mind, everything must be expressed in terms of ft, sec, and lb. Given the velocity 25 mph, we can't say much before converting it to ft/sec! Let's do that:

V = 25 mph * 5280 feet/mile / 3600 (sec/hour)

V = 36.67 ft/sec

Now that we have a velocity in terms of ft/sec, we can substitute values for every term in the equation except for x:

`1/2(9450)x^2 = 4000/(2(32.2))(36.67)^2`

Simplifying:

`4725x^2 = 83521`

Divide by 4725:

`x^2 = 17.68`

Take the square root of both sides:

`x = 4.20`

Therefore, we need the spring to be able to compress a minimum of 4.20 ft.

﻿Hope that helps!