Video Transcript
A force 𝐅 equal to negative four
𝐢 minus nine 𝐣 newtons is acting on a particle whose position vector as a function
of time is given by 𝐫 of 𝑡 is equal to negative nine 𝑡 minus eight 𝐢 plus
negative three 𝑡 squared plus two 𝐣 meters. Calculate the work 𝑤 done by the
force 𝐅 between 𝑡 equals three and 𝑡 equals eight seconds.
In this question, we are asked to
find the work done by a force. This is equal to the scalar or dot
product of the force vector 𝐅 and displacement vector 𝐝. We are told that the force acting
on the particle is equal to negative four 𝐢 minus nine 𝐣 newtons. We are also given the position
vector as a function of time and need to calculate the work done between 𝑡 equals
three and 𝑡 equals eight seconds. When 𝑡 is equal to three, the
position vector 𝐫 of 𝑡 is equal to negative nine multiplied by three minus eight
𝐢 plus negative three multiplied by three squared plus two 𝐣. This simplifies to negative 35𝐢
minus 25𝐣.
When 𝑡 is equal to eight seconds,
the position vector of the particle is negative nine multiplied by eight minus eight
𝐢 plus negative three multiplied by eight squared plus two 𝐣. This is equal to negative 80𝐢
minus 190𝐣. We are now in a position to
calculate the displacement over this time period. We subtract the position vector at
𝑡 equals three from the position vector at 𝑡 equals eight. Collecting the 𝐢- and
𝐣-components gives us negative 45𝐢 minus 165𝐣.
We now have the force vector in
newtons and displacement vector in meters. As already mentioned, the work done
is the dot or scalar product of these. And this is equal to the sum of the
products of the individual components. We have negative four multiplied by
negative 45 plus negative nine multiplied by negative 165. This is equal to 180 plus 1485
which is equal to 1665. As we are working in standard
units, the work done by the force 𝐅 between 𝑡 equals three and 𝑡 equals eight
seconds is 1665 joules.
