Two tugboats pull a disabled super-tanker. Each tug exerts a constant force of 2.1×106 N , one an angle 19 ∘ west of north and the other an angle 19 ∘ east of north, as they pull the tanker a distance 0.60 km toward the north. What is the total work they do on the super-tanker?

2.38 x 10⁹ Nm

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

1 - wave length

2 - amplitude

3 - crest

4 - trough

explanation:

duh.