Work and Energy Theorem- change in acceleration

[JhonnyO]

1. The problem statement, all variables and given/known data

A toy car of mass 555g is moving with speed 1.21m/s and strikes a spring mounted to the wall. The spring slows the toy car to rest with a acceleration that varies from 0 - 3.055m/s2.
Determine the spring constant.

2. Relevant equations
W = -1/2mv2 + 1/2 kx2




3. The attempt at a solution
I've seen equations like this but I don't understand the acceleration part. I think that because there is no friction the work is conservative so it equals zero. Then I solved the equation for x2 and substituted that back into the equation so that I could solve for k but the answer I got doesn't jive with the one my instructor gave me. What am I doing wrong?


Thanks for your help.

[rl.bhat]

W = -1/2mva + 1/2 kx2
It should be W = -1/2mva^2 + 1/2 kx2

[JhonnyO]

Why is the acceleration part of the kinematic equation? And if there is a range (0-3.055) then what number should I insert?

Otherwise was the way I was going about solving it right?

Thanks

[rl.bhat]

For kinetic energy you have used 1/2* m *va. It should be 1/2* m *va^2. Is it typo?

[JhonnyO]

oh yea, sorry but the acceleration should be included?

[LowlyPion]

No. a is not a part of the equation. I think there is some confusion in your notation, since I trust you didn't intend that a was a subscript of v.

KE = 1/2*m*v2

[JhonnyO]

my question is still how does the acceleration factor into this problem? I'm trying to work the problem out and I'm not getting the question right I don't understand what I am doing wrong.

am I using the right formula?

[LowlyPion]

You have 2 equations and 2 unknowns don't you?

You don't know k or x.

But they tell you that max a = 3.055 which means that

Fmax = .555*3.055

which is also

Fmax = k*x

You also know that

1/2*m*v2 = 1/2*k*x2

So Xmax = m*v2/Fmax

and k = Fmax/Xmax