Q6E - University Physics with Modern Physics

Question

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?

Step-by-Step Solution

Answer

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

Step 1: Identification of the given data

The given data is listed below as-

• The magnitude of the force acting on the tanker is, $F_{tugboat 1}=F_{tugboat 2}=1.8\times {10}^{6}N$ 
• The angle between the force and the two tugboats is, ${\varphi }_1=\varphi =14°$
• The displacement, s=0.75 km 

Step 2: Significance of the work done on a particle by a constant force

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 $\varphi =180°$. 

Step 3: Determination of work done by tugboats on the supertanker

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 is the 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 $2.62\times {10}^{9}J$ .