Question

A sailboat moves north for a distance of 10.00 km when blown by a wind 30° east of south with a force of 5.00×10^4 N . How much work was done by the wind?

Answer 1

Answer:

-433 MJ of work

Explanation:

Given:

Displacement of the sailboat is, $d=10.00\ km$ towards north

Force applied by the wind is, $F_w=5.00\times 10^4\ N$

Direction of the force is, $\theta=30(Towards\ East\ of\ South)$

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:

$W=F\cdot d=Fd\cos x$

Where, 'x' is the angle between the tails of the vectors 'F' and 'd'.

Now, from the figure below, we can find 'x'.

$x=180-\theta=180-30=150$

Now, plug in all the given values and solve for 'W'.

$W=(5.00\times 10^4\ N)(10.00\times 10^3\ m)(\cos 150)\\\\W=-433012702\ J =-433\ MJ$

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.