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

Answer:

Car body rise on its suspension by 0.0309 m

Explanation:

We have given mass of the car m = 1500 kg

Mass of each person = 90 kg

Speed of the car $v=20km/hr=20\times \frac{5}{18}=5.555m/sec$

Distance traveled by car d = 3.7 m

So time period  $T=\frac{distance}{speed}=\frac{4}{5.55}=0.72sec$

Frequency $f=\frac{1}{T}=\frac{1}{0.72}=1.388Hz$

Angular frequency is $\omega =2\pi f=2\times 3.14\times 1.388=8.722rad/sec$

Angular frequency is equal to $\omega =\sqrt{\frac{k}{m}}$

$8.722 =\sqrt{\frac{k}{1500}}$

k = 114109.92 N/m

Now weight of total persons will be equal to spring force

$4mg=kx$

$4\times 90\times 9.8=114109.92\times x$

x = 0.0309 m