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PhoenixFire Group Title

A horizontal spring with constant k=1200 and mass 10kg attached to the end. The mass is on a horizontal frictionless surface. Question: How much power is required to compress the spring 70mm in a period of 2 seconds?

  • 2 years ago
  • 2 years ago

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  1. akash123 Group Title
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    how much work one will do in compressing the spring by 70 mm?

    • 2 years ago
  2. akash123 Group Title
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    or what's the energy stored in the spring when it's compressed by 70 mm?

    • 2 years ago
  3. PhoenixFire Group Title
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    \[E_f={1 \over 2}kx^2={1 \over 2}(1200)(0.07^2)=2.94 [J]\] \[E_i=0\] To find the power it's this equation right? \[P={{\Delta E} \over {\Delta t}}\]

    • 2 years ago
  4. akash123 Group Title
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    yes.....good:)

    • 2 years ago
  5. PhoenixFire Group Title
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    \[P={2.94-0 \over 2}=1.47\] What are the units of Power?

    • 2 years ago
  6. akash123 Group Title
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    watt or joule/s

    • 2 years ago
  7. PhoenixFire Group Title
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    Oh right... because it's Energy over time... energy is Joule, time is seconds.. duh. Thanks!

    • 2 years ago
  8. akash123 Group Title
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    yes...:)

    • 2 years ago
  9. PhoenixFire Group Title
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    To find the initial acceleration if the spring was released do i use F=ma? Calculate the Force using Hooke's Law \[F=-kx\] and plug into \[a={F \over m}\] Right? But is x negative as it's compressed by 70mm or is it positive?

    • 2 years ago
  10. akash123 Group Title
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    but we make some assumptions in this question i.e. we are compressing the spring smoothly(slowly) such that there is no change in kinetic energy of mass attached with the spring.

    • 2 years ago
  11. akash123 Group Title
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    x will be negative since spring is under compression

    • 2 years ago
  12. PhoenixFire Group Title
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    So then F=-kx=-1200(-0.07)=84 a=F/m=84/10=8.4 ms-2

    • 2 years ago
  13. PhoenixFire Group Title
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    I think they're trying to trick us. Because if that's right above it doesn't factor in friction, and the next question asks what if the surface has friction of 0.65 what is the initial acceleration. Well it would be the same answer.

    • 2 years ago
  14. akash123 Group Title
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    no...acceleration of the block wont be same if there is friction..|dw:1347418241566:dw|

    • 2 years ago
  15. PhoenixFire Group Title
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    Ahhhhh so it would be the net force acting on the block divided by it's mass. Which in this case would be: \[a={F_{net} \over m}={{F_{fric}+F_{spring}} \over m}\]

    • 2 years ago
  16. akash123 Group Title
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    so net force on the block= ( spring force- frictional force)

    • 2 years ago
  17. PhoenixFire Group Title
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    Negative frictional force because it's acting in the opposite direction. \[a={{F_{spring}-F_{fric}} \over m}={{-kx-\mu mg} \over m}={{84-63.765} \over 10}=2.0235 [ms^{-2}]\] That's a big difference.

    • 2 years ago
  18. akash123 Group Title
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    yes...but how did u get the fictional force?

    • 2 years ago
  19. PhoenixFire Group Title
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    \[\mu mg=(0.65)(10)(9.81)=63.765\]

    • 2 years ago
  20. akash123 Group Title
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    ok...you have done correctly.

    • 2 years ago
  21. PhoenixFire Group Title
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    Awesome, Thank you @akash123 This question is complete.

    • 2 years ago
  22. akash123 Group Title
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    :)

    • 2 years ago
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