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

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

Step-by-step 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 = (m * g )/(x )`

=>
`k = (68 * 9.8)/( 0.012)`

=>
`k = 55533.33 \ N/m`