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A worker drives a .500kg spike into a rail tie with a 2.50kg sledgehammer. The hammer hits the spike with a speed of 65.0 m/s. If one-third of the hammer's kenetic energy is converted to the internal energy of the hammer and spike, how much does the total internal energy increase?

User Sphinx
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Answer:

The increase internal energy is 1,760.41
\overline 6 J

Step-by-step explanation:

The given parameters of the worker, sledgehammer, and rail line are;

The mass of the spike driven into the rail, m₁ = 0.500 kg

The mass of the sledgehammer, m₂ = 2.50 kg

The speed of the sledgehammer, v = 65.0 m/s

The fraction of the hammers kinetic energy, KE, converted to the internal energy of the hammer and the spike, Δ
E_(internal) = 1/3 × KE

Therefore, we have;

The kinetic energy, KE, of the sledgehammer is given by KE = 1/2·m₂·v²

Where;

m₂ = The mass of the sledge hammer = 65.0 m/s

v = The velocity o the sledgehammer

∴ KE = 1/2 × 2.50 kg × (65.0 m/s)² = 5,281.25 J

Therefore;

The increase internal energy, Δ
E_(internal) = 1/3 × 5,281.25 J = 1760.41
\overline 6 J

User Nicholas Cardot
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