A 1175 kg car carrying four 82 kg people travels over a rough "washboard" dirt road with corrugations 4.0 m apart which causes the car to bounce on its spring suspension. The car bounces with maximum amplitude when its speed is 24 km/h. The car now stops, and the four people get out. By how much does the car body rise on its suspension owing to this decrease in weight?

To find out how much the car body rises on its suspension owing to the decrease in weight, we need to determine the change in the weight of the car after the four people get out.

First, let's calculate the initial weight of the car using the given information that the car carries four 82 kg people. The weight of each person can be found by multiplying their mass by the acceleration due to gravity (g).

Weight of each person = mass × acceleration due to gravity

Weight of each person = (82 kg) × (9.8 m/s²) = 803.6 N (rounded to one decimal place)

Total weight of the people = weight of each person × number of people

Total weight of the people = (803.6 N) × 4 = 3214.4 N (rounded to one decimal place)

Now, let's calculate the initial weight of the car by adding the weight of the people to the weight of the car itself.

Initial weight of the car = weight of the people + weight of the car

Given that the car weighs 1175 kg, the weight of the car can be found using the same formula:

Weight of the car = mass × acceleration due to gravity

Weight of the car = (1175 kg) × (9.8 m/s²) = 11515 N (rounded to one decimal place)

Initial weight of the car = Weight of the people + Weight of the car

Initial weight of the car = 3214.4 N + 11515 N = 14729.4 N (rounded to one decimal place)

After the four people get out, the weight of the car decreases by the weight of the people.

Change in weight of the car = Weight of the people

Change in weight of the car = 3214.4 N

Therefore, the car body rises on its suspension by 3214.4 N, owing to the decrease in weight after the four people get out.

I will be happy to critique your thinking.