Question

When three people with a total mass of 2.00 x 102 kg step into their 1.200 x 103 kg car, the car’s

springs are compressed by 3.0 cm.

1.2.a. What is the spring constant of the car’s springs assuming they act as a single spring?
1.2.b. How far will the car lower if loaded with 3.00 x 102 kg rather than 2.00 x 102 kg​

Answer 1

a

`k = 457333.3 N/m`

b

`x_a =0.09\ m`

Step-by-step explanation:

From the question we are told that

The total mass of three people is
`M = 2.00*10^(2) \ kg`

The mass of the car is
`m_c = 1.200 *10^(3) \ kg`

The compression of the car spring is
`x = 3 \ cm = 0.03 \ m`

Generally the spring constant is mathematically represented as

`k = (F)/(x)`

Here F is the force exerted by the mass of three people and that of the car , this is mathematically represented as

=>
`F = (M +m_c) *g`

=>
`F = ([2.0*10^(2) ]+[ 1.200*10^(3)]) * 9.8`

=>
`F = 13720 \ N`

So

`k = (13720)/(0.03)`

=>
`k = 457333.3 N/m`

Generally if the mass which the car is loaded with is
`m = 3.00*10^(2) \ kg`

Then the force experienced by the spring is

=>
`F_a = (m +m_c) *g`

=>
`F_a = (3.00*10^(3) + 1.200 *10^(3)) * 9.8`

=>
`F_a = 41160 \ N`

Generally from the above formula the compression is

`x_a = (F_a)/(k)`

=>
`x_a = (41160)/(457333.3)`

=>
`x_a =0.09\ m`