Question

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= ____________

Answer 1

Answer:

The approximate  spring constant is  $k = 55533.33 \ N/m$

Explanation:

From the question we are told that

   The  mass of the person is  $m = 68 \ kg$

     The  dip of the car is  $x = 1.2 \ cm = 0.012 \ m$

Generally according to hooks law  

        $F = k * x$

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

         $F = m * g$

=>    $m * g = k * x$

=>     $k = \frac{m * g }{x }$

=>    $k = \frac{68 * 9.8}{ 0.012}$

=>   $k = 55533.33 \ N/m$