A fisherman is fishing from a bridge and is using a "49.0-N test line." In other words, the line will sustain a maximum force of 49.0 N without breaking. What is the weight of the heaviest fish that can be pulled up vertically, when the line is reeled in (a) at constant speed and (b) with an acceleration whose magnitude is 2.35 m/s2?

To find the weight of the heaviest fish that can be pulled up vertically, we need to consider the tension in the fishing line. The tension in the line is equal to the weight of the fish.

(a) When reeling in the line at a constant speed, the net force is zero since there is no acceleration. Therefore, the tension in the line is equal to the weight of the fish. Using the formula for weight, we can calculate the weight of the fish:

Weight of the fish = Tension in the line = 49.0 N

So, the weight of the heaviest fish that can be pulled up vertically at a constant speed is 49.0 N.

(b) When reeling in the line with an acceleration of 2.35 m/s^2, the net force acting on the fish is the difference between the tension in the line and the weight of the fish. The net force is given by:

Net force = Tension in the line - Weight of the fish

Since the fish is being pulled up vertically, its weight is given by the formula:

Weight of the fish = mass of the fish × acceleration due to gravity

We can rearrange this equation to solve for the mass of the fish:

Mass of the fish = Weight of the fish / acceleration due to gravity

The acceleration due to gravity is approximately 9.8 m/s^2. Substituting this value into the equation, we get:

Mass of the fish = Weight of the fish / 9.8

Now, let's substitute this expression for mass into the equation for net force:

Net force = Tension in the line - (Weight of the fish / 9.8)

Since the net force is equal to the mass of the fish multiplied by the acceleration, we have:

Mass of the fish × 2.35 = Tension in the line - (Weight of the fish / 9.8)

Simplifying the equation, we get:

2.35 × Mass of the fish = Tension in the line - (Weight of the fish / 9.8)

Multiplying through by 9.8 to remove the denominator, we obtain:

2.35 × 9.8 × Mass of the fish = 9.8 × Tension in the line - Weight of the fish

Since the tension in the line is equal to the weight of the fish, we can substitute the weight of the fish for the tension:

2.35 × 9.8 × Mass of the fish = 9.8 × Weight of the fish - Weight of the fish

Simplifying further:

2.35 × 9.8 × Mass of the fish = 9.8 × Weight of the fish - Weight of the fish

20.3 × Mass of the fish = 8.8 × Weight of the fish

Dividing both sides by the weight of the fish (which is equivalent to the tension in the line), we obtain:

20.3 × Mass of the fish / Weight of the fish = 8.8

Now, we can substitute the value of the tension in the line (49.0 N) for the weight of the fish:

20.3 × Mass of the fish / 49.0 = 8.8

Rearranging the equation to solve for the mass of the fish:

Mass of the fish = (8.8 × 49.0) / 20.3

Calculating this expression, we find:

Mass of the fish ≈ 21.36 kg

So, the weight of the heaviest fish that can be pulled up vertically with an acceleration of 2.35 m/s^2 is approximately 21.36 kg.

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