Solving Hooke's Law Problem: Spring Constant & Speed of 370kg Car

[Abdul.119]

Homework Statement

You have been hired to design a spring-launched roller coaster that will carry two passengers per car. The car goes up a 10-m-high hill, then descends 20m to the track's lowest point. You've determined that the spring can be compressed a maximum of 2.5m and that a loaded car will have a maximum mass of 450kg . For safety reasons, the spring constant should be 12% larger than the minimum needed for the car to just make it over the top.

What spring constant should you specify?
What is the maximum speed of a 370kg car if the spring is compressed the full amount?

Homework Equations

(1/2) kx^2 = mgh
(1/2)kx^2 = (1/2) mv^2

The Attempt at a Solution

I solved the first question which was 1.6*10^4 (rounded to two sig figs) and the system accepted it.. but it won't accept my answer for the second one, please see if I really have anything wrong:

(1/2)kx^2 = (1/2) mv^2

v = sqrt(kx^2/m)

v = sqrt(1.6*10^4 N/m * (2.5 m)^2 / 450 kg)

v = 14.8 m/s

To two sig figs = 15 m/s.. which it says is wrong..

Thanks

 

[Chestermiller]

You calculated the velocity when it loses contact with the spring. They want the velocity after it goes down the 20 meters to the track's lowest point (probably given that its velocity is essentially zero at the top of the hill).

 

[nasu]

And they want the speed for a 370 kg car. You used 450 kg, didn't you?