Fx = 5.75 cos 57.5
Fy = 5.75 sin 57.5
Fz = 0
F = Fx i + Fy j + 0 k
D= 4.8 i - 2.2 j + 1.7 k
F dot D = (5.75 cos 57.5) * 4.8 - (5.75 sin 57.5 * 2.2) + 0 * 1.7 Joules
A constant force of magnitude 5.75 N is exerted on an object. The force's direction is 57.5° counterclockwise from the positive x axis in the xy plane, and the object's displacement is
Δr = (4.8i − 2.2j + 1.7k) m.
Calculate the work done by this force.
J
2 answers
xy plane, but directions are i,j. Goodness, you are headed for a breakdown. Lets work in the i,j plane
work=force dot distance
force=5.75cos57.5deg i +5.75sin57.5deg j=3.09i + 4.85j
work= (3.09i + 4.85j). (4.8i − 2.2j + 1.7k) =3.09*4.8+4.85*(-2.2) + 0= you do it.
check my math, typing math is not my thing.
work=force dot distance
force=5.75cos57.5deg i +5.75sin57.5deg j=3.09i + 4.85j
work= (3.09i + 4.85j). (4.8i − 2.2j + 1.7k) =3.09*4.8+4.85*(-2.2) + 0= you do it.
check my math, typing math is not my thing.
4 answers