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The four wheels of a car are connected to the car's body by spring assemblies that let the wheels move up and down over bumps and dips in the road. When a 68 kg (about 150 lb) person sits on the left front fender of a small car, this corner of the car dips by about 1.2 cm (about 12 in). If we treat the spring assembly as a single spring, what is the approximate spring constant?

Respuesta :

Answer:[tex]k=55.590 KN/m[/tex]

Explanation:

Given

mass of person[tex]\left ( m\right )[/tex]=68 kg

car dips about 1.2 cm

We know

F=kx

Where k=combined  spring constant

mg=kx

[tex]k=\frac{mg}{x}[/tex]

[tex]k=\frac{68\times 9.81}{1.2\time 10^{-2}}[/tex]

[tex]k=55.590 KN/m[/tex]

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