Video Transcript
A particle moves in a plane in
which 𝐢 and 𝐣 are perpendicular unit vectors. Its displacement from the origin at
time 𝑡 seconds is given by 𝐫 equals two 𝑡 squared plus seven 𝐢 plus 𝑡 plus
seven 𝐣 meters. And it is acted on by a force 𝐅
equals six 𝐢 plus three 𝐣 newtons. How much work does the force do
between 𝑡 equals two seconds and 𝑡 equals three seconds?
Since work done is the dot product
of the vector force and the vector for displacement, we’re going to need to
calculate the displacement of this particle. We’re given its displacement at
some time 𝑡. But we’re also asked how much work
the force does between 𝑡 equals two seconds and 𝑡 equals three seconds. So, let’s substitute 𝑡 equals two
and 𝑡 equals three into our expressions for the displacement. Beginning with 𝑡 equals three
seconds, we get two times three squared plus seven 𝐢 plus three plus seven 𝐣. That’s 25𝐢 plus 10𝐣. And we’re working in meters. So the displacement from the origin
and, hence, the position of the particle at three seconds is 25𝐢 plus 10𝐣
meters. In a similar way, we can find the
position of the particle at two seconds by substituting 𝑡 equals two into our
expression. And we get 15𝐢 plus nine 𝐣
meters.
The displacement then of the
particle between these two times is the difference here. It’s 𝐝 sub three minus 𝐝 sub
two. That’s 25𝐢 plus 10𝐣 minus 15𝐢
plus nine 𝐣. And of course, we simply subtract
the individual components. 25 minus 15 is 10, and 10 minus
nine is one. So, the displacement is simply 10𝐢
plus 𝐣, and that’s in meters. Since the work done is the dot
product of the force and displacement, we need to find the dot product of six 𝐢
plus three 𝐣 and the displacement we just calculated. It’s the dot or scalar product of
six 𝐢 plus three 𝐣 and 10𝐢 plus 𝐣. That’s six times 10 plus three
times one, which is, of course, equal to 63. The work done then is equal to 63
joules.
