91Ó°ÊÓ

The given data is listed below as-

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

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

$$\begin{gathered} \mathit{W}\mathbf{=}\stackrel{\mathbf{→}}{\mathbf{F}}.\stackrel{\mathbf{→}}{\mathit{S}} \\ \mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{=}\mathit{F}\mathit{s}\mathbf{cos}\mathbf{}\mathit{ϕ} \end{gathered}$$

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

The total work done by tugboats on the supertanker will be

$W_{total}=\left(F_{tugboat1}×s×\cos \mathrm{ϕ}\right)+\left(F_{tugboat2}×s×\cos \mathrm{ϕ}\right)$

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

$W_{total}=2\left(F×s×\cos \mathrm{ϕ}\right)$

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

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

$W_{total}=2\left(F×s×\cos \mathrm{ϕ}\right)$

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

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