How to find the acute angle that a constant unit force vector makes with the positive x-axis if the work done by the force in moving a particle from (0,0) to (4,0) equals 2.

Question

anonymous · Answer

work = Fx x = displacement in force direction = √((0-0)^2 + (4-0)^2) = 4 units in the positive x direction 2 = F * 4 F = force = 2/4 = 1/2 N (the horizontal component of the force) F(horizontal) = F * cos Φ cosΦ = F(horizontal) / F = (1/2) / 1 = 1/2 Φ = arc cos 1/2 = 60 Source: https://answers.yahoo.com/question/index?=20130609074425AA336K6

irishboy123 · Answer

or, for constant force, $W = \vec F \bullet \vec x = |\vec F|| \vec x| \cos \theta = 2$
here $|\vec F|\ = 1$ and $|\vec x|\ = 4$