Question #9510

a 2.44 kg sledgehammer hits the spike with a speed of 63.6 m/s. how much does the total internal energy increase?

Expert's answer

dW=mv22\mathrm{d}W = \frac{mv^2}{2}dW=1.22×63.62=4934.8\mathrm{dW} = 1.22 \times 63.6^2 = 4934.8

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