The striking part of a hammer whose weight is 10 kg freely falls from a height of 2.3 m onto an iron cart weighing

The striking part of a hammer whose weight is 10 kg freely falls from a height of 2.3 m onto an iron cart weighing 150 kg. How many blows did the hammer make if the cart was heated by 20 K? Heating consumes 30% of the hammer’s energy. Specific heat capacity of iron C = 460 J \ kg * K g = 10 mg (s * s)

To determine the number of blows made by the used hammer, we use the formula: η = Ap / Az = Czh * mp * Δt / (n * mm * g * h), whence we express: n = Czh * mp * Δt / (η * mm * g * h).

Variables and constants: Szh – specific heat capacity of iron (Szh = 460 J / (kg * K)); mp is the mass of the cart (mp = 150 kg); Δt is the change in the temperature of the carriage (Δt = 20 K); η – process efficiency (η = 30% = 0.3); mm is the mass of the striking part of the hammer (mm = 10 kg); g – acceleration of gravity (g = 10 m / s2); h – the height of the fall of the hammer (h = 2.3 m).

Calculation: n = Czh * mp * Δt / (η * mm * g * h) = 460 * 150 * 20 / (0.3 * 10 * 10 * 2.3) = 20 * 10 ^ 3 strokes.

Answer: The hammer made 20 * 10 ^ 3 hits on the cart.