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Question on work energy and power kinetic energy of an ....

  1. Aug 18, 2016 #1
    1. The problem statement, all variables and given/known data
    If the speed of a vehicle increases by 2m/s then it's kinetic energy is doubled. Find out the original speed of the vehicle

    2. Relevant equations
    1/2(mv^2)

    3. The attempt at a solution
    First find out both KE(kinetic energy) taking u(initial vel.)
    And v(final vel.) And mass as m
    K(initial KE)= 1/2(mu^2)
    K`(final KE)=1/2(mv^2)
    =1/2[m(u+2)^2]. {Given "speed of vehicle increases by 2m/s}

    Also
    2K=K`
    Then equating
    2[1/2(mu^2)]=1/2[m(u+2)^2].
    =>mu^2=1/2(mu^2+4m+4mu)
    =>u^2=1/2(u^2+4+4u) {dividing m on both sides}
    =>2u^2=u^2+4+4u
    =>0= -u^2+4+4u.
    {Further using quadratic formula}
    We get
    u=(√2+1)÷2
    Sadly that ain't there in the options so. :~( {booo}
     
  2. jcsd
  3. Aug 18, 2016 #2

    SammyS

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    Check your work in using the quadratic formula.
     
  4. Aug 18, 2016 #3
    Will do
     
  5. Aug 18, 2016 #4
    OK redid it made a slight mistake but it's still no good

    u=(-1+/-√2)÷2
     
  6. Aug 18, 2016 #5
    The +/- means plus or minus or this ± sorry didn't notice the sign before
     
  7. Aug 18, 2016 #6

    SammyS

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    That's also wrong.

    Show your steps.

    (The initial try looked better.)
     
  8. Aug 18, 2016 #7

    Doc Al

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    Your quadratic looks good, but redo your application of the quadratic formula.

    (Edit: SammyS beat me to it!)
     
  9. Aug 18, 2016 #8
    OK got the mistake the ans now is 2(√2+1)
    Which is correct I was taking the denominator wrong thank you very much for pointing that out
     
  10. Aug 18, 2016 #9
    Thank you all for helping
     
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