To solve this problem we will start by applying the equilibrium equations, for which the net force applied on the body is given by the difference between the weight of the object and the tension applied on the fishing line. Starting from finding the force, we can proceed to find the maximum resistant acceleration of the object and finally applying the linear kinematic equations for the duration of the process. Now calculating the Force we have,
[tex]F_{net} = T-mg[/tex]
[tex]F_{net} = 60N - (5.0kg)(9.81m/s^2)[/tex]
[tex]F_{net} = 11N[/tex]
By Newton's second law we have finally that,
[tex]F = ma[/tex]
Here m is the mass and a the acceleration,
[tex]11N = (5.0kg)(a)[/tex]
[tex]a = 2.2m/s^2[/tex]
The time by the fish can be calculated through the kinematic linear equation, then
[tex]x = v_0t + \frac{1}{2} at^2[/tex]
There is not initial velocity because the fish initially is at rest,
[tex]x = \frac{1}{2} at^2[/tex]
Rearranging to find the time,
[tex]t = \sqrt{\frac{2x}{a}}[/tex]
[tex]t = \sqrt{\frac{2(2.0m)}{2.2m/s^2}}[/tex]
[tex]t = 1.35s[/tex]
Therefore the amount of time in which the fisherman can raise the fish to the dock without losing it is 1.35s
