A 3.20kg fish is attached to the lower end of a vertical spring that has negligible mass and force constant 905N/m . The spring initially is neither stretched nor compressed. The fish is released from rest.

1- What is its speed after it has descended 0.0460m from its initial position?Express your answer with the appropriate units.

2- What is the maximum speed of the fish as it descends?Express your answer with the appropriate units.

Explanation:

1) Let its velocity be v after it descends by .046 m

loss of PE  = gain of  spring energy + 1/2 mv²

3.2 x 9.8 x .046 = 1/2 x 905 x .046² + 1/2 x 3.2 x v²

= 1.4425 = .9575 + 1.6 v²

v = .55 m / s

2 )

At maximum speed , net force = 0

mg = kx

x = mg / k

= 3.2 x 9.8 / 905

= .0346 m

Speed will be maximum when the spring is stretched by .0346 m

Using the formula stated above we can calculate velocity . That is

loss of PE  = gain of  spring energy + 1/2 mv²

3.2 x 9.8 x .0346 = 1/2 x 905 x .0346² + 1/2 x 3.2 v²

1.0851 = .5417 +1.6 v²

v = .5827 m /s

the answer is d. 4

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