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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 1/2 in).

If we treat the spring assembly as a single spring, what is the approximate spring constant?

k= ____________

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okpalawalter8

Answer:

The approximate  spring constant is  [tex]k = 55533.33 \ N/m[/tex]

Explanation:

From the question we are told that

   The  mass of the person is  [tex]m = 68 \ kg[/tex]

     The  dip of the car is  [tex]x = 1.2 \ cm = 0.012 \ m[/tex]

Generally according to hooks law  

        [tex]F = k * x[/tex]

here the force F is the weight of the person which is mathematically represented as

         [tex]F = m * g[/tex]

=>    [tex]m * g = k * x[/tex]

=>     [tex]k = \frac{m * g }{x }[/tex]

=>    [tex]k = \frac{68 * 9.8}{ 0.012}[/tex]

=>   [tex]k = 55533.33 \ N/m[/tex]

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