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dainel40 Group Title

The first figure gives spring force Fx versus position x for the spring–block arrangement of the second figure. The scale is set by Fs = 190 N. We release the block at x = 16.0 cm. How much work does the spring do on the block when the block moves from xi =+8.0 cm to (a)x=+6.0 cm, (b)x=-6.0 cm, (c)x=-8.0 cm, and (d)x=-10.0 cm?

  • one year ago
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  1. dainel40 Group Title
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    • one year ago
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    • one year ago
  3. azolotor Group Title
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    Hooke's Law is \[F = k \delta x\] where k is the spring constant and delta x is the displacement of the spring. Work is defined as \[W =Fdcos(\theta)\] where theta is the angle between the two vectors when they are drawn tail to tail. So, first we calculate the force generated by the displacement of the spring and then from there since the force is parallel to the distance the block is traveling we get \[W=Fd\] so once the force is found you can calculate the work at each displacement.

    • one year ago
  4. dainel40 Group Title
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    yes but I'm stuck I can calculate the force but I dont know what to do after. Work is force*distance

    • one year ago
  5. azolotor Group Title
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    They give you the distances so if you have the force you can calculate the work just plug into the above formula

    • one year ago
  6. dainel40 Group Title
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    but wouldnt I have to subtract from my original distance x=16?

    • one year ago
  7. azolotor Group Title
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    If I am reading this correctly no because they are asking for the distance from x(i). If I misread this then yes you just need to determine the actual displacement that's all.

    • one year ago
  8. dainel40 Group Title
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    okay thanks!

    • one year ago
  9. dainel40 Group Title
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    yea that didn't work just by pluging it in

    • one year ago
  10. dainel40 Group Title
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    and I tried subtracting the distance and I got it wrong too

    • one year ago
  11. azolotor Group Title
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    let met take a look

    • one year ago
  12. dainel40 Group Title
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    thanks

    • one year ago
  13. azolotor Group Title
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    First of all make sure you are putting the right sign on everything. Also you can extrapolate from the graph since it is a line you can easily find the equation for it and find the values at the respective x values to calculate the work. Also, make sure you convert to SI units

    • one year ago
  14. dainel40 Group Title
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    Could u do step A as an examples then I will follow ur steps for the rest

    • one year ago
  15. azolotor Group Title
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    Sure hold on

    • one year ago
  16. azolotor Group Title
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    The graph has a slope of -1 and you can pick the point 0,0 for ease since it passes through the origin so y=-x. It wants it at +8cm so that is y=-8 so 8 newtons in the negative direction of force is exerted. It went a distance of 0.08 m so 0.08m *8N =64 Joules

    • one year ago
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