A 2.4 kg iron ball is dropped from a height of 14 m onto a concrete roadway, and

Question

A 2.4 kg iron ball is dropped from a height of 14 m onto a concrete roadway, and 2.5% of its kinetic energy at the time it reaches the ground is transformed into internal energy in the ball itself. (The rest of the energy is transmitted to the ground, converted into sound energy, and so on.) What is the ball's increase in temperature?

Explanation / Answer

v^2 = 0 + 2*(9.8)*(14) => v = 16.565 m/s => Kinetic energy = (1/2)*2.4*(16.565)^2 = 329.28 J => change in internal energy = (2.5/100)*329.28 = 8.232 J since, specific heat of iron = 0.45 J/Kg/K => 2.4*0.45*T = 8.232 => T = 7.6 C => cchange in temperature = 7.6 degrees..

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