The spring constant of an automobile suspension spring increases with increasing

Question

The spring constant of an automobile suspension spring increases with increasing load due to a spring coil that is widest at the bottom, smoothly tapering to a smaller diameter near the top. The result is a softer ride on normal road surfaces from the wider coils, but the car does not bottom out on bumps because when the lower coils collapse, the stiffer coils near the top absorb the load. For such springs, the force exerted by the spring can be empirically found to be given by F = axb.For a tapered spiral spring that compresses 12.4 cm with a 1000 N load and  33cm with a 5000 N load, do the following.
(a) Evaluate the constants a and b in the empirical equation F.
a = ? N/mb
b = 1.66?

(b) Find the work needed to compress the spring 24.5 cm.
=? J

I m not really sure how it wants me to find a and b. I would really appreciate some help.

Explanation / Answer

F = -kx
1000 = a * 0.124^b ...............(1)
5000 = a * 0.330^b ................(2)

From (1):
a = 1000/0.124^b

Sub into (2):

5000 = (1000/0.124^b) * 0.330^b = (330 / 0.124)^b
ln 5000 = b ln (330 / 0.124)
b = ln 5000 / (ln 330 - ln 0.124)
b = ln 5000 /7.886

b = 1.64

so

a = 1000/0.124^(1.64)

a = 30675.25 N/mb

b)

W = ab(1/2)x

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