Question

A 2.85 kg fish is attached to the lower end of a vertical spring that has negligible mass and force constant 875 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 0.0580 m from its initial position? B)What is the maximum speed of the fish as it descends?

Answer 1

Answer:

A) V = 0.324 m/s

B) V = 0.56 m/s

Explanation:

A) Change in gravitational potential energy = mass * gravity * height change

Change in G.P.E = 2.85 * 9.81 * 0.058 = 1.6215 J

Energy transferred to spring = 0.5 * k * x^2

Here k = 875 N/m

And x = 0.058 m

Thus energy transferred to spring = 1.47175 J

The G.P.E is converted to spring potential energy and kinetic energy as it moves down. So the difference in energy is accounted by kinetic energy as follows:

Kinetic energy = G.P.E - Spring P.E

Kinetic energy = 1.6215 - 1.47175 = 0.14975 J

We can now find the speed using this kinetic energy:

Kinetic energy = 0.14975

0.5 * mass * velocity^2 = 0.14975

0.5 * 2.85 * V^2 = 0.14975

V = 0.324 m/s

B) The fish accelerates because the force on it are unbalanced. These forces are the weight of the fish, and the force of the spring stopping it.

As long as the weight of the fish is more than the upward force of the spring, the fish will continue to accelerate. Using this knowledge, we can deduce that the speed is maximum when the weight and spring force are equal. Thus we set them equal and find out the displacement first:

Weight = spring force

2.85 * 9.81 = 875 * Displacement

Displacement = 0.03195 m

Similarly as (A):

Change in G.P.E = 2.85 * 9.81 * 0.03195 = 0.8933 J

Spring P.E = 0.5 * 875 * 0.03195^2 = 0.4466 J

Kinetic energy = 0.8933 - 0.4466 = 0.4467 J

0.5 * mass * V^2 = 0.4467

0.5 * 2.85 * V^2 = 0.4467

V = 0.56 m/s

Answer 2

(a) The speed of the spring when it descends 0.058 m is 1.033 m/s.

(b) The maximum speed of the fish is 0.032 m.

The given parameters;

• mass of the fish, m = 2.85 kg
• spring constant, k = 875 N/m

The speed of the spring when it descends 0.058 m is calculated as follows;

$\frac{1}{2}mv^2 = \frac{1}{2} kx^2\\\\mv^2 = kx^2\\\\v^2 = \frac{kx^2}{m} \\\\v = \sqrt{\frac{kx^2}{m} } \\\\v = \sqrt{\frac{875 \times 0.058^2}{2.85} }\\\\v = 1.033 \ m/s$

The maximum speed of the fish is calculated as follows;

$mg = kx\\\\x = \frac{mg}{k} \\\\x = \frac{2.85 \times 9.8}{875} \\\\x = 0.032 \ m$

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