physics

A 10 kg block slides up a hill to a height of 5.8 m. If 414 J of thermal energy are generated, how fast was the block going at the bottom of the hill?
1 m/s

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asked by ami
  1. Potential energy at the top of the hill
    = mgh
    = 10*9.81*5.8 J
    =569 J
    Heat energy produced
    = 414 J

    If we assume conservation of energies, energy at the bottom of the hill in the form of kinetic energy
    =(1/2)mv² = 569+414 =983

    Solve for v:
    v=sqrt(2*983/10)
    =14.0 m/s

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    posted by MathMate

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