A person fishing hooks a 3.0-kg fish on a line that can only sustain a maximum of 35 N of force before breaking. At one point while reeling in the bass, it fights back with a force of 45 N. What is the minimum acceleration with which you must play out the line during this time in order to keep the line from breaking?

If the fish does actually exert a force of 45 N on the line, it will break. It exerts its force mainly on the water around it, away from the fisherman. If it accelerates backwards in the process due to released line, a line at its breaking point exerts a force on the fish (and vice versa) equal to 45 N - Ma = 35 N. The backwards acceleration of the fish is then

a = 10 N/g = 1.02 m/s^2

To find the minimum acceleration with which you must play out the line during this time, we can use Newton's second law of motion, which states that the force applied to an object is equal to the mass of the object multiplied by its acceleration (F = ma).

In this case, the force applied to the line is equal to the tension in the line. We can represent this force as T. The mass of the fish is given as 3.0 kg, and the maximum tension the line can sustain before breaking is 35 N.

Let's assume the tension in the line when the fish fights back is T_fish. This tension must be less than or equal to 35 N to prevent the line from breaking. Therefore, we can write an equation: T_fish ≤ 35 N.

Now, let's consider the force exerted by the fish. The force exerted by the fish is given as 45 N. According to Newton's second law, this force is equal to the product of the mass of the fish (3.0 kg) and its acceleration (a_fish): 45 N = (3.0 kg) * a_fish.

Since we want to find the minimum acceleration with which you must play out the line to prevent it from breaking, we need to find the acceleration of the fish. Rearranging the equation, we get: a_fish = 45 N / 3.0 kg.

Now, we know that the force exerted by the fish is equal to the tension in the line (T_fish). So, we can write: T_fish = (3.0 kg) * a_fish.

Substituting the value of a_fish from the previous equation, we have: T_fish = (3.0 kg) * (45 N / 3.0 kg).

Simplifying, T_fish = 45 N.

Therefore, the minimum acceleration with which you must play out the line during this time to prevent it from breaking is zero, since the tension in the line when the fish fights back is already less than or equal to the maximum tension the line can sustain before breaking (35 N).