Asail boat moves north for a distance of 10 km when blown by a wind 30° east of south with the force of 5.00×10^4 n. how much work was done by the wind?
a) 5.00 x 10^8 j
b) 2.50 x 10^8 j
c) 0.00 j
d) 4.33 x 10^8 j

a) 5.00 x 10^8 J

The work done to move the sailboat is calculated through the equation;

W = F x d

where F is force and d is the distance.

Substituting the known values from the given above,

                             W = (5.00 x 10⁴ N)(10 km)(1000 m/ 1km)

                                 = 5.00 x 10⁸ J

Thus, the work done is 5.00 x 10⁸Joules

the electric field is a “force field” around a charged object that illustrates the direction the electric force would push an imaginary