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A fisherman is fishing from a bridge and is using a "40.0-N test line." In other words, the line will sustain a maximum force of 40.0 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 2.57 m/s2?

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To solve this problem we require the concept of Forces given by Newton in his second law. For the particular case where the acceleration is 0 and when there is a change of acceleration.

A) For the first case where the speed is constant we have to

[tex]F = mg[/tex]

Here,

[tex]a=0m/s^2[/tex]

Then,

[tex]F= mg[/tex]

[tex]40N = m (9.8)[/tex]

[tex]m = 4.077kg[/tex]

B) Here we have an acceleration of [tex]2.57m / s ^ 2[/tex]. When there is acceleration it is considered that the sum of forces is dynamic, for that reason when performing the summation, we have

[tex]\sum F = ma[/tex]

[tex]F-mg = ma[/tex]

[tex]T = m(a+g)[/tex]

[tex]40= m(9.8+2.57)[/tex]

[tex]m = 3.23kG[/tex]

Therefore the heaviest fish that can be pulled up vertically is 3.23KG when there is acceleration but 4.077kg with constant velocity

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