What is the force needed to stretch a horizontal spring by 7.15 cm?

[destro47]

One end of a horizontal spring (k = 333 N/m) is attached to a 3.12 kg box, and the other end to a fixed, vertical wall. (A picture of this situation can be found in Fig 4-10 on page 102 of your textbook.)

a) Find the magnitude of the force needed to move the mass so the spring is stretched by 7.15 cm.

b) Suppose the spring is stretched 7.15 cm and released. When the spring is stretched by only 2.88 cm, what is the acceleration of the mass?

Tried it plugging in the given quantities into hookes equation, doesn't work out that way. How do I approach this one? Here is what I'm thinking so far:

F=ma
where F in this case is the same F in hookes equation, thus:

-kx= ma


Am I on the right track here?

 

[blochwave]

You should be thinking in terms of energy, remember the potential energy of a spring displaced from equilibrium by a distance of x is 1/2*k*x^2

Also knowing that W=F*d, you can solve a)

Similarly for b) you start with all potential energy and at x=2.88 you have part potential and part kinetic, kinetic of course involves the speed, which you can solve for

EDIT: For a) I think I just solved for the force required to stretch the spring and you'd also have to add to that expression for w, so W=Wspring+Wmass, so the work done just displacing the spring + the work done just displacing the mass. I think

 

[destro47]

Sorry but my class has not covered W yet. I believe its W=fd, right? My professor is a real ball buster who genuinely enjoys making his students squirm over the homework assignments. Any other suggestions?