Question

A 3.10 kg fish is attached to the lower end of a vertical spring that has negligible mass and force constant 850 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 0.0460 m from its initial position?

b)What is the maximum speed of the fish as it descends?

Answer 1

a)

By Conservation of energy

mgx =(1/2)mV²+(1/2)Kx²

3.1*9.81*0.046 =(1/2)*3.1*V² + (1/2)*850*0.046²

V=0.5677m/s

b)

Maximum Speed occurs when ,dV/dx=0

By Conservation of energy

mgx =(1/2)mV²+(1/2)Kx²

Differentiating w.r.t x

mg =kx+m(dV/dx)

mg=kx

=>x=mg/k =3.1*9.8/850 =0.03574 m

3.1*9.8*0.03574 =(1/2)*850*0.03574² +(1/2)*3.1*V²

V=0.5924 m/s