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A 72 kg meteorite buries itself 5.0 m into soft mud. The force between the meteorite and the mud is given by x^3, where x is the depth in the mud. What was the speed of the meteorite when it first hit the mud?
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Added Sun, 21 Jun '15
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The energy absorbed by the earth is

W = integral of [F(x)*dx]

W = 660* integral of [x^3*dx][x = 0 to 5]

W = 660*[(1/4)*x^4][x = 0 to 5]

W = 660/4*5^4 = 103125 J

This must equal the kinetic energy of the meteorite when it impacts, so

0.5*m*V² = 103125

V = sqrt[2*103125/72] = 52.79 m/s
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Added Sun, 21 Jun '15
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