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Energy Question Problem

  1. Nov 23, 2013 #1
    A cyclist is competing in a race and decides to pass some fellow cyclists who are getting tired. The initial speed of the cyclist is 21 km/h and he has 3.0 * 10^2 W of power. If the bicycle is able to convert 92% of input energy into kinetic energy, how fast will cyclist be travelling after 4.0 s. The combined mass of the cyclist and the bicycle is 78 kg.

    The answer is 7.9 m/s. But I dont know how.
     
  2. jcsd
  3. Nov 23, 2013 #2
    show us your attempt.... then we will help you...!!!
     
  4. Nov 23, 2013 #3
    energy= Pt= 3x10^2 * 4 = 1.2 x10^3 J
    Useful energy = 1.2 x10^3 *0.92 = 1.104 x10^3
    KE= 1/2m( v)^2
    root(1.104 x10^3 *2 /78) = change in v = +/-5.32 m/s (3 S.F.). He is accelerating hence +5.32m/s
    21km/h = 21000m/60x60s= 5.83 m/s (3 S.F)
    Total speed = 5.32+5.83 = 11.15m/s
     
  5. Nov 23, 2013 #4
    useful energy =1.104x10^3 ... right
    then change in kinetic energy will be equal to useful energy given in 4 seconds.......
    the formula 1/2mv^2, 'V' is the instataneous velocity, not the change in velocity
    so we get
    change in kinetic energy = useful energy given..
    mx(Vfinal2-Vinitial2)/2
     
  6. Nov 23, 2013 #5
    I dont understand your solution. Can you please Clarify
     
  7. Nov 23, 2013 #6

    Dick

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    Whoa, back it up a bit. You've got the energy input correct. What's your initial kinetic energy? Add the energy input to that to get the final kinetic energy. Use that to get a final velocity.
     
  8. Nov 23, 2013 #7
    initial kinetic energy + useful energy given= final kinetic energy
     
  9. Nov 23, 2013 #8
    hope this will make you clear
     
  10. Nov 23, 2013 #9
    I get 21 which is not the correct? Please Help
     
  11. Nov 23, 2013 #10
    never mind i got it.

    Thanks alot
     
  12. Nov 23, 2013 #11
    he he he ...........
    if you have really got it then it is fine, if not we are always to help you.....!!
     
  13. Nov 23, 2013 #12

    Vanadium 50

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