Answer to Question #82803 in Mechanics | Relativity for Luz

Question

A force F~ = Fx ˆı + Fy ˆ acts on a particle that
undergoes a displacement of ~s = sx ˆı + sy ˆ
where Fx = 8 N, Fy = −2 N, sx = 6 m, and
sy = 3 m.
Find the work done by the force on the
particle.
Answer in units of J.

Answer 1

F=F_x+F_y=8i-2j=8.25e^(-14.04j)

s=s_x+s_y=6i+3j=6.71e^26.57j

A=F·s=8.25e^(-14.04j)·6.71e^26.57j=55.36e^12.53j