A 3.00 kg fish is attached to the lower of a vertical spring that has negligible

Question

A 3.00 kg fish is attached to the lower of a vertical spring that has negligible mass and force constant 900 N/m. The spring initially is neither stretched nor compressed. The fish is released from rest. (a) What is its speed after it has descended 0.0500m from its initial position? (B) What is the maximum speed of the fish as it descends?

Explanation / Answer

we have to deal with three forms of energy: potential energy = mgs kinetic energy = 1/2 mv^2 spring energy = 1/2 ks^2 at the rest point is the potential energy relative to the position 0.05 m lower = mgs = 3*9.81*0.05 J kinetic energy = 0 spring energy = 0 after releasing and 0.05 m lower is the potential energy = 0 kinetic energy = 1/2 *3*v^2 J spring energy = 1/2 * 900*0.05^2 and Epot = Ekin + Espr 3*9.81*0.05 = 1/2 *3*v^2 + 1/2 * 900*0.05^2 --> v = 0.480625 m/s (speed after 5 cm) ------------------ going back to the formula for v, we see that v = v(2(mgs - 1/2*ks^2)/m) v = v(2gs - ks^2/m) v = v(2*9.81s - 300s^2) now we look for the maximum of this function: the derivative is v' = (9.81 - 300s)/v(s(19.62-300s)) = 0 so 9.81 = 300s s = 327/10 000 (3.27 cm) and v (max) = 327v3/1000 = 0.566381 m/s (max speed )

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