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Posted by on Friday, April 27, 2012 at 8:39pm.

A 50.0kg cyclist on a 10.0kg bicycle speeds up from 5.0m/s to 10.0m/s , the total kinetic energy before acceleratingis 750, and after acceleratingis 3000, how much work is done to increase the kinetic energy of the cyclist and the bicycle

  • physics - , Saturday, April 28, 2012 at 11:13am

    Work for changing KE of cyclist+bicycle is
    W=Δ KE= KE2 –KE1=
    =3000-750 = 2250 J,
    for the cyclist:
    W1=ΔKE1 = m1•v2^2/2 – m1•v1^2/2 =
    =50•100/2-50•25/2 =1875 J,
    for the bicycle:
    W2=ΔKE2 = m2•v2^2/2 – m2•v1^2/2 =
    =10•100/2 - 10•25/2 =375 J

  • physics - , Wednesday, January 27, 2016 at 1:27pm

    I am 4 years in the future

  • physics - , Tuesday, February 16, 2016 at 1:15am

    dude same

  • physics - , Monday, March 28, 2016 at 10:55am

    me too

  • physics - , Sunday, May 1, 2016 at 4:15pm

    We could also use:

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