The tension at which a fishing line snaps is commonly called the line’s “strength.” What minimum strength is needed for a line that is to stop a salmon of weight 85 Nin 11 cmif the fish is initially drifting at$\mathbf{2}\mathbf{.}\mathbf{8}\mathbf{}\mathbf{m}\mathbf{/}\mathbf{s}$? Assume a constant deceleration.

$$\begin{gathered} 1)\mathrm{Initial}\mathrm{velocity}\mathrm{is},{\mathrm{V}}_0=2.8\mathrm{m}/\mathrm{s} \\ 2)\mathrm{The}\mathrm{final}\mathrm{velocity}\mathrm{is},\mathrm{v}=0\mathrm{m}/\mathrm{s} \\ 3)\mathrm{The}\mathrm{displacement}\mathrm{is}∆\mathrm{x}=0.11\mathrm{m} \\ 4)\mathrm{Mass}\mathrm{of}\mathrm{salmon}\mathrm{is}\mathrm{m}=8.7\mathrm{kg} \end{gathered}$$

The force acting on the object is equal to the product of mass and acceleration of the object. The acceleration is the rate of change of velocity with respect to time.

Assume that the fishing line and the path of the salmon are horizontal. If we assume, the direction of the salmon’s velocity along the positive x-axis, its acceleration (it is deceleration) would be negative, as the negative force acts upon it to stop a salmon. So, the force of tension is in the negative direction of the x-axis.

Formulae:

$$\begin{gathered} {\mathrm{V}}_{\mathrm{f}}^2={\mathrm{V}}_0^2+2\mathrm{a}∆\mathrm{X}\left(1\right) \\ \mathrm{F}=\mathrm{ma}\left(2\right) \end{gathered}$$

To find the acceleration, use equation (i) and substitute the value of initial and final velocity.

$$\begin{gathered} {\mathrm{V}}^2={\mathrm{V}}_0^2+2\mathrm{a}∆\mathrm{x} \\ \mathrm{a}=-\frac{{\mathrm{V}}_0^2}{2∆\mathrm{x}} \end{gathered}$$

The initial velocity is given and the final velocity is zero as the salmon finally stops. Therefore,

$$\begin{gathered} \mathrm{a}=\frac{2.8\mathrm{m}/\mathrm{s}}{2\times \left(0.11\mathrm{s}\right)} \\ =-36\mathrm{m}/{\mathrm{s}}^2 \end{gathered}$$

Acceleration is negative because the stopping force is against the direction of motion.

The magnitude of the acceleration is$36\mathrm{m}/{\mathrm{s}}^2$

The strength required is equal to the tension in the line. The tension in the line is equivalent to the force. To calculate the force, use the equation (ii).

T=F

=ma

Substitute the value of mass and acceleration.

$$\begin{gathered} \mathrm{T}=\left(8.7\mathrm{kg}\right)\left(36\mathrm{m}/{\mathrm{s}}^2\right) \\ =3.1\times {10}^2\mathrm{N} \end{gathered}$$

Hence, the minimum strength required is$3.1\times {10}^{2}N$.