Asked by Moe

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?
Answered by Elena
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 = …
Answered by Moe
You saved my life Elena. Much appreciated.
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