Question

A force F~ = Fxˆı + Fyˆ acts on an object with Fx = 4 N and Fy = 6 N. The angle between F~ and the displacement vector ~s is 18◦ , and 106 J of work is done by F~. Find the magnitude of ~s. Answer in units of m

Answer 1

The magnitude of the vector displacement is 15.65 m.

Resultant force

The resultant force acting on the object is calculated as follows;

$F = \sqrt{F_x^2 + F_y^2} \\\\F = \sqrt{4^2 + 6^2} \\\\F = 7.21 \ N$

Displacement of the vector

The displacement of the vector is calculated as follows;

W = Fs cosθ

$s = \frac{W}{Fcos(\theta)} \\\\s = \frac{106}{7.12 \times cos(18)} \\\\s = 15.65 \ m$

Learn more about work done by vector here: https://brainly.com/question/25573309