Answer:
2.38 x 10⁹ Nm
Step-by-step explanation:
To calculate the total work done by the forces from the tugboats, we first find their work done by individual force.
Let the;
work done by the first tugboat be W₁ ,
constant force exerted by first tugboat be F₁ = 2.1 x 10⁶N
which is inclined 19° west of north
and which moves the super-tanker a distance, D = 0.6km north = 600m;
And;
work done by the second tugboat be W₂
constant force exerted by second tugboat be F₂ = 2.1 x 10⁶N
which is inclined 19° east of north
and which moves the super-tanker a distance, D = 0.6km north = 600m;
These can be represented by the following figure;
D
North
F₁ \ | / F₂
\ | /
\ | /
West \|/ East
|
|
|
|
South
Now, we know that the work done (W) by a force (F) during a displacement (D) is given by the dot product of F and D as follows;
=> W = F . D -------------------------[Where F and D are vectors]
=> W = F x D cos θ; ---------------------(i)
Where;
F = magnitude of vector F
D = magnitude of vector D
θ = angle between F and D
(A) Using equation (i), we can write;
W₁ = F₁ x D cos θ₁ -------------------(ii)
Where;
θ₁ = angle between F₁ and D = 19°
Substituting the values of F₁ , D and θ₁ into equation (ii) gives;
W₁ = 2.1 x 10⁶ x 600 x cos 19°
W₁ = 2.1 x 10⁶ x 600 x 0.9455
W₁ = 1.19 x 10⁹ Nm
(B) Also using equation (i), we can write;
W₂ = F₂ x D cos θ₂ -------------------(iii)
Where;
θ₂ = angle between F₂ and D = 19°
Substituting the values of F₂ , D and θ₂ into equation (iii) gives;
W₂ = 2.1 x 10⁶ x 600 x cos 19°
W₂ = 2.1 x 10⁶ x 600 x 0.9455
W₂ = 1.19 x 10⁹ Nm
(C) Therefore the total work done (W) is the sum of W₁ and W₂ as follows;
W = W₁ + W₂
W = (1.19 x 10⁹) + (1.19 x 10⁹)
W = 2.38 x 10⁹ Nm
The total work done on the super-tanker is 2.38 x 10⁹ Nm