Solve Hooke's Law Problem: Spring Constant & Compression

[MozAngeles]

Homework Statement

A 60 kg driver gets into an empty taptap to start the day's work. The springs compress 0.02 m. What is the effective spring constant of the spring system in the taptap?
using F=kx I got 2.94*10^4
Then
After driving a portion of the route, the taptap is fully loaded with a total of 25 people with an average mass of 60 kg per person. In addition, there are three 15 kg goats, five 3 kg chickens, and a total of 25 kg of bananas on their way to the market. Assume that the springs have somehow not yet compressed to their maximum amount. How much are the springs compressed?
I used the formula F=kx, solving for x, but i got the wrong answer (1.80*10^4) the answers turns out to be .5328. I just don't know how to get there.
For F i just multiplied each set of things times the number there was of them times the mass (kg) times 9.8 m/s^2 to get Newtong for each, then i divided that by the constant i found previously.

Homework Equations

The Attempt at a Solution

I used the formula F=kx, solving for x, but i got the wrong answer (1.80*10^4) the answers turns out to be .5328. I just don't know how to get there.
For F i just multiplied each set of things times the number there was of them times the mass (kg) times 9.8 m/s^2 to get Newtong for each, then i divided that by the constant i found previously.

 

[zorro]

The work done by man in compressing the spring is equal to the gain in elastic energy of the spring.

mgx=0.5kx^2
solve this and you will get value of k
Rest is done using same equation.

 

[MozAngeles]

i realized all i was doing wrong was the total mass of the taptap thanks anyways..