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torisob [31]
3 years ago
13

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?
Physics
1 answer:
Westkost [7]3 years ago
4 0

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

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Explanation:

Hi there!

The average speed (a.s) is calculated by dividing the traveled distance (d) over the time needed to travel that distance (t):

a.s = d / t

So, let´s find the distance traveled in those 3 s. For that, we can use the equation of position of an object moving in a straight line with constant acceleration:

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x0 = initial position.

v0 = initial velocity.

t = time.

a = acceleration.

If we place the origin of the frame of reference at the point where the runner starts, then, x0 = 0. Since the runner starts from rest, v0 = 0. So, the equation gets reduced to this:

x = 1/2 · a · t²

We have the time (3 s), so let´s find the acceleration. For that, we can use the equation of velocity of an object moving in a straight line with constant acceleration:

v = v0 + a · t

Where "v" is the velocity at a time "t". Since v0 = 0, then:

v = a · t

At t = 3 s, v = 8 m/s

8 m/s = a · 3 s

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So let´s find the position of the runner at t = 3 s (In this case, the position of the runner will be equal to the traveled distance):

x = 1/2 · a · t²

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Now, we can calcualte the average speed:

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The average speed of the runner is 4 m/s.

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