Answer:
[tex]W_t = 3.125*10^9 J[/tex]
Explanation:
We know that the work could be calculate dby:
W = [tex]F_xd[/tex]
where [tex]F_x[/tex] is the component of the force in the direction of the
displacement. and d is the displacement.
So, the total work ([tex]W_t[/tex]) is:
[tex]W_t = F_1cos(\beta) d+F_2cos(\beta)d[/tex]
Where [tex]F_1[/tex] is the force done by the first tugboat, [tex]\beta[/tex] is the angle of the force, d is the displacement and [tex]F_2[/tex] is the force done by the second tugboat.
[tex]W_t = (1.9*10^6)cos(19°)(870m)+(1.9*10^6)cos(19°)(870m)[/tex]
[tex]W_t = 3.125*10^9 J[/tex]
This question involves the concepts of vector components and the work done.
The total work done by the two tugboats on the supertanker is "3.12 x 10⁹ J".
First, we will calculate the force in the direction of the displacment. It will be the sum of the x- vector components of both the forces:
[tex]F_{x}=F_{x1}+F_{x2}\\F_{x}=F\ Cos\ \theta_1+F\ Cos\ \theta_2[/tex]
where,
F = magnitude of forces = 1.9 x 10⁶ N
θ₁ = angle of the first boat with horizontal = 19°
θ₂ = angle of the second boat with horizontal = 19°
Therefore,
[tex]F_x=2(1.9\ x\ 10^6\ N)(Cos\ 19^o)\\F_x=3.59\ x\ 10^6\ N[/tex]
Now, we will calculate the work done by the tugboats:
[tex]W = F_xd\\W=(3.59\ x\ 10^6\ N)(870\ m)[/tex]
W = 3.12 x 10⁹ J
Learn more about work here:
brainly.com/question/13662169?referrer=searchResults
The attached picture explains the work done formula.
