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Work and Energy Theorem- change in acceleration

  • Thread starter JhonnyO
  • Start date
  • #1
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Homework Statement



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.

Homework Equations


W = -1/2mv2 + 1/2 kx2




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.
 
Last edited:

Answers and Replies

  • #2
rl.bhat
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W = -1/2mva + 1/2 kx2
It should be W = -1/2mva^2 + 1/2 kx2
 
  • #3
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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
 
  • #4
rl.bhat
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For kinetic energy you have used 1/2* m *va. It should be 1/2* m *va^2. Is it typo?
 
  • #5
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oh yea, sorry but the acceleration should be included?
 
  • #6
LowlyPion
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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
 
  • #7
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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?
 
  • #8
LowlyPion
Homework Helper
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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
 
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