Answer:
-433 MJ of work
Explanation:
Given:
Displacement of the sailboat is, [tex]d=10.00\ km[/tex] towards north
Force applied by the wind is, [tex]F_w=5.00\times 10^4\ N[/tex]
Direction of the force is, [tex]\theta=30(Towards\ East\ of\ South)[/tex]
The vector diagram representing the given scenario is shown below.
We know that, work done by a force is the dot product of force and displacement and is given as:
[tex]W=F\cdot d=Fd\cos x[/tex]
Where, 'x' is the angle between the tails of the vectors 'F' and 'd'.
Now, from the figure below, we can find 'x'.
[tex]x=180-\theta=180-30=150[/tex]
Now, plug in all the given values and solve for 'W'.
[tex]W=(5.00\times 10^4\ N)(10.00\times 10^3\ m)(\cos 150)\\\\W=-433012702\ J =-433\ MJ[/tex]
Therefore, the work done by the wind is nearly 433 MJ. The negative sign implies that the force acts in the direction opposite to the displacement.
