Two tugboats pull a disabled tanker. Each tug exerts a constant force of 1.80x10⁶ N one$\mathbf{14}\mathbf{°}$west of north and the other$\mathbf{14}\mathbf{°}$east of north, as they pull the tanker 0.75 km toward the north. What is the total work they do on the supertanker?

The given data is listed below as-

• The magnitude of the force acting on the tanker is, $F_{tugboat1}=F_{tugboat2}=1.8\times {10}^{6}N$
• The angle between the force and the two tugboats is,${\varphi }_1=\varphi =14°$
• The displacement, s=0.75 km

Work done on a particle by a constant force$\overrightarrow{\mathbf{F}}$during a linear displacement$\overrightarrow{\mathbf{s}}$ is given by

$$\begin{gathered} \mathit{W}\mathbf{=}\overrightarrow{\mathbf{F}}.\overrightarrow{\mathit{S}} \\ \mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{=}\mathit{F}\mathit{s}\mathbf{cos}\mathbf{}\mathit{\varphi } \end{gathered}$$

If F and s are in the same direction then $\varphi =0$ and if it is in opposite direction then $\varphi =180°$.

The total work done by tugboats on the supertanker will be

$W_{total}=\left(F_{tugboat1}\times s\times \cos \varphi \right)+\left(F_{tugboat2}\times s\times \cos \varphi \right)$

The magnitude of force applied by the two tugboats and also the angles are equal. Therefore,

$W_{total}=2\left(F\times s\times \cos \varphi \right)$

Here, F isthe magnitude of the force acting on the tanker, s is the displacement,

For $F_{tugboat1}=F_{tugboat2}=1.8\times {10}^{6}N$, s=0.75 km and $\varphi =14°$

$W_{total}=2\left(F\times s\times \cos \varphi \right)$

$$\begin{gathered} W_{total}=2\times \left(1.8\times {10}^{6}N\right)\times 0.75\times {10}^{3}m\times \left(\cos 14°\right) \\ =2.62\times {10}^{9}N.m \\ =2.62\times {10}^{9}J \end{gathered}$$

Thus, the total work done by the tugboats on the supertanker is$2.62\times {10}^{9}J$.