Acar with a total mass of 1800 kg (including passengers) is driving down a washboard road with bumps spaced 4.9 m apart. the ride is roughest—that is, the car bounces up and down with the maximum amplitude—when the car is traveling at 5.7 m/s. what is the spring constant of the car's springs? express your answer to two significant figures and include the appropriate units.

k = 9.6 x 10^5 N/m or 9.6 kN/m

Explanation:

First, we need to use the expression to calculate the spring constant which is:

w² = k/m

Solving for k:

k = w²*m

To get the angular velocity:

w = 2πf

The problem is giving the linear velocity of the car which is 5.7 m/s. With this we can calculate the frequency of the car:

f = V/x

f = 5.7 / 4.9 = 1.16 Hz

Now the angular velocity:

w = 2π*1.16

w = 7.29 rad/s

Finally, solving for k:

k = (7.29)² * 1800

k = 95,659.38 N/m

In two significant figures it'll ve 9.6 kN/m

i would say