The mass of ice melted as a result of friction between the ice and the horizontal surface is 2.78g
<u>Explanation:</u>
Given,
Temperature, T = 0°C
Initial mass, Mi = 62kg
Speed, s = 5.48m/s
Distance, x = 26.8m
Friction is present.
Mass of ice melted = ?
We know,
The amount of energy required for the melting of ice is exactly equal to the initial kinetic energy of the block of ice
and

Therefore, 
KE = 930.94 Joules
Ice melting lateral heat is 334 kJ/kg = 334000 J/kg.
Therefore, the melted mass of the ice = 930.94 / 334000 = 0.00278 kg = 2.78 g.
Thus, The mass of ice melted as a result of friction between the ice and the horizontal surface is 2.78g
1 watt = 1 joule/second
1 horsepower = 746 watts = 746 joule/second
(150 horsepower) x (746 watt/HP) x (1 joule/sec / watt) x (10 sec)
= (150 x 746 x 1 x 10) joule = 1,119,000 joules .
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