Posted by **AE** on Friday, December 4, 2009 at 8:35pm.

An 8.56 kg block of ice at 0 degrees celsius is sliding on a rough horizontal icehouse floor (also at 0 degrees celsius) at 15.6 m/s. Assume that half of any heat generated goes into the floor and the rest goes into the ice. How much ice (in kg) has melted after the speed of the ice has been reduced to 10.2 m/s? What is the maximum amount of ice that will melt?

- physics -
**MathMate**, Friday, December 4, 2009 at 9:00pm
Use energy considerations.

m=8.56 kg

v0=15.6 m/s,

v1=10.2 m/s

latent heat of fusion of ice

= 333.55 KJ/kg

Initial kinetic energy

=(1/2)mv0²

Final kinetic energy

=(1/2)mv1²

Half of energy lost goes to melt the ice block

Ef=(1/2)m(v0²-v1²)

=(1/2)8.56(15.6²-10.2²)

= 596.3 J

Amount of melted ice

= ((596.3/2)/1000)KJ /333.55 (KJ/kg)

= 0.9 g

Maximum amount of ice would melt if all the kinetic loss goes to melting the ice.

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