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Posted by on Monday, January 28, 2013 at 3:10pm.

A solid gold bathtub (mass=250 Kg)is pushed along a horizontal cement road (mew=0.59)by criminals. The criminals, in fear of capture, run away and leave the bathtub sliding with a velocity of 9 m/s. Assuming all kinetic energy is converted to heat from frictional work. 1.How far does the golden tub slide?
2.What is the change in the tubs temperature if the specific heat of gold is 129 J/Kg C?
3.How much energy would you have to add to increase the temperature of the bath tub from 25 degrees Celsius to 27 degrees Celsius?

  • Physics - , Monday, January 28, 2013 at 3:35pm

    1.
    mv²/2=μmgs
    s=v²/2μg
    2.
    mv²/2=Q=cmΔT
    ΔT=v²/2c
    (c=129000 J/kg•C)
    3.
    Q=cmΔT= mv²/2
    Q1= cmΔT1=cm•2=…
    ΔQ=Q1-Q= cm•2- mv²/2 = …

  • Physics - , Monday, January 28, 2013 at 3:37pm

    You saved my life Elena. Much appreciated.

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