Video Transcript
A body moves in a plane in which 𝐢
and 𝐣 are perpendicular unit vectors. Two forces 𝐅 sub one equals nine
𝐢 minus two 𝐣 newtons and 𝐅 sub two equals nine 𝐢 minus seven 𝐣 newtons act on
the body. The particle moves from the point
with position vector negative six 𝐢 plus two 𝐣 meters to the point two 𝐢 plus
three 𝐣 meters. Find the work done by the resultant
of the forces.
We begin by recalling that we can
find the work done by a force vector 𝐅 over some displacement vector 𝐝 by finding
the dot product of these two vectors. We do need to be a bit careful here
though, because we’re working with two forces and we want to find the work done by
their resultant. Now, of course, the resultant of
two forces is simply their sum. So, the resultant force 𝐅 is the
sum of 𝐅 sub one and 𝐅 sub two. In this case, that’s nine 𝐢 minus
two 𝐣 plus nine 𝐢 minus seven 𝐣. Now, of course, we simply add the
components, so we’ll add the 𝐢-components first. Nine plus nine is 18. Then, we add the 𝐣-components. Negative two minus seven is
negative nine. And so, the resultant of the forces
in this case is 18𝐢 minus nine 𝐣 newtons.
Now, we need to find the
displacement. Now, of course, the particle moves
from the point with position vector negative six 𝐢 plus two 𝐣 to the point two 𝐢
plus three 𝐣. Let’s call the first point 𝐴 and
the second point 𝐵. Then, the displacement is the
vector 𝐀𝐁. Well, that’s the difference between
vector 𝐎𝐁 and 𝐎𝐀. So, it’s two 𝐢 plus three 𝐣 minus
negative six 𝐢 plus two 𝐣. As before, we simply subtract the
individual components, beginning with the 𝐢-components. Two minus negative six is eight,
and three minus two is one. So, we find the displacement of the
body is eight 𝐢 plus 𝐣. And of course, that’s in
meters.
And we have everything we need. The work done is simply the dot
product or the scalar product of these two vectors. That’s 18 times eight plus negative
nine times one, which is 144 minus nine, or 135. The work done by the resultant of
the two forces then is 135 joules.
