Question

A 1500 kg car carrying four 90 kg people travels over a "washboard" dirt road with corrugations 3.7 m apart. The car bounces with maximum amplitude when its speed is 20 km/h. When the car stops, and the people get out, by how much does the car body rise on its suspension?

Answer 1

Car body rise on its suspension by 0.0309 m

Step-by-step explanation:

We have given mass of the car m = 1500 kg

Mass of each person = 90 kg

Speed of the car
`v=20km/hr=20* (5)/(18)=5.555m/sec`

Distance traveled by car d = 3.7 m

So time period
`T=(distance)/(speed)=(4)/(5.55)=0.72sec`

Frequency
`f=(1)/(T)=(1)/(0.72)=1.388Hz`

Angular frequency is
`\omega =2\pi f=2* 3.14* 1.388=8.722rad/sec`

Angular frequency is equal to
`\omega =\sqrt{(k)/(m)}`

`8.722 =\sqrt{(k)/(1500)}`

k = 114109.92 N/m

Now weight of total persons will be equal to spring force

`4mg=kx`

`4* 90* 9.8=114109.92* x`

x = 0.0309 m