Question

car rides on four wheels that are connected to the body of the car by spring assemblies that let the wheels move up and down over bumps and dips in the road. When an 80 kg person sits on the left front fender of a small car, this corner of the car dips by about 1 cm. Treating the spring assembly as a single spring, what is the approximate spring constant

Answer 1

Answer:

approximate spring constant = 78.5 KN/m

Explanation:

From the question,

Mass of person = 80kg

Dip of car = 1cm

From Hooke's law, we know that;

F = kx = mg

Where ;

k is combined spring constant

x = 1cm = 0.01m

g = 9.81 m/s²

Thus,

Kx = mg

K = mg/x

k = (80x9.81)/0.01

k = 78.48 KN/m

This is approximately 78.5 KN/m:

Answer 2

Answer:

78.4 KN/m

Explanation:

Given

mass of person 'm' =80 kg

car dips about i.e spring stretched 'x'=   1 cm  => 0.01m

acceleration due to gravity 'g'= 9.8 m/s^2

as we know that,in order to find approximate spring constant we use Hooke's Law i.e  F=kx

where,

F = the force needed

x= distance the spring is stretched or compressed beyond its natural length

k= constant of proportionality called the spring constant.

F=kx ---> (since f=mg)

mg=kx

k=(mg)/x

k=(80 x 9.8)/ 0.01

k=78.4x$10^3$

k=78.4 KN/m