A steam hammer weighing 4 tons falls on the anvil from a height of 2.5 m. Determine the force of the hammer’s impact on the workpiece if the impact lasted 0.01 s. Will the impact force of the hammer change if it jumped to a height of 30 cm upon impact?
m = 4 t = 4000 kg.
h1 = 2.5 m.
h2 = 30 cm = 0.3 m.
t = 0.01 s.
g = 9.8 m / s ^ 2.
m * g * h1 = m * V1 ^ 2/2.
The speed of the hammer V at the moment of the start of the blow will be determined by the formula: V1 = √ (2 * g * h1).
The speed of the hammer V1 “at the end of the blow will be equal to 0: V1” = 0.
Let’s write 2 Newton’s law: m * (V1 – V1 “) / t = F1.
F1 = m * V1 / t = m * √ (2 * g * h1) / t
F1 = 4000 kg * √ (2 * 9.8 m / s ^ 2 * 2.5 m) / 0.01 s = 2800000 N.
The speed of the hammer V2 “at the end of the blow for the second case will be equal: V2 = √ (2 * g * h2) and directed up.
Let’s write 2 Newton’s law: m * (V1 – (- V2 “)) / t = F2.
F2 = m * (V1 + V2 “) / t = m * (√ (2 * g * h1) + √ (2 * g * h2)) / t = m * √ (2 * g * (h1 + h2) ) / t.
F2 = 4000 kg * √ (2 * 9.8 m / s ^ 2 * (2.5 m + 0.3 m)) / 0.01 s = 2963241 N.
Answer: F1 = 2800000 N, F2 = 2963241 N.
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