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?

The increase internal energy is 1,760.41 J

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 = 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 = 1/3 × 5,281.25 J = 1760.41 J

radiation

explanation:

heat is transmitted by radiation