Question

Car enthusiasts often lower their cars closer to the ground as a matter of style. James wants to lower his car by replacing all four of his original coil springs with new ones offered as an option by the car manufacturer. Note that changing the coil springs on a car can unsafely affect its handling. Consult your national or state laws before altering the suspension on any car. The new springs will be identical to the original springs, except the force constant will be 5355.00 N/m smaller. When James removes the original springs, he discovers that the length of each spring expands from 8.55 cm (its length when installed) to 12.00 cm (its length with no load placed on it). If the mass of the car body is 1455.00 kg, by how much will the body be lowered with the new springs installed, compared to its original height?

Answer 1

Answer:

$\Delta h=0.0364\ m=3.64\ cm$

Explanation:

Given:

• change in stiffness constant of the spring on replacing the original springs, $\delta k=5355\ N.m^{-1}$

• mass of the car, $m=1455\ kg$

• initial length of the original car-spring before compression, $l_i=12\ cm=0.12\ m$

• final length of the original car-spring after compression, $l_f=8.55\ cm=0.0855\ m$

So, weight of the car:

$w=m.g$

$w=1455\times 9.81$

$w=14273.55\ N$

Now the spring constant of original spring:

$w=4k_o.(l_i-l_f)$ (since 4 springs are in parallel)

$14273.55=4k_o\times (0.12-0.0855)$

$k_o=103431.522\ N.m^{-1}$

So the stiffness constant of the new springs:

$k_n=k_o-\delta k$

$k_n=103431.522-5355$

$k_n=98076.522\ N.m^{-1}$

Now the height lowered:

$w=k_n.4\Delta h$ (since 4 springs are in parallel)

$14273.55=4\times98076.522\times \Delta h$

$\Delta h=0.0364\ m=3.64\ cm$