Video Transcript
A particle moved from point 𝐴
negative two, negative two to point 𝐵 six, 10 along a straight line under the
action of the force 𝐅 equal to 𝑘𝐢 minus six 𝐣 acting in the opposite direction
to the displacement 𝐀𝐁. Find the work done by the force
𝐅.
We begin by recalling that the work
done by a force can be calculated by finding the dot product of the vector force 𝐅
and displacement force 𝐝. In this question, we are told that
𝐅 is equal to 𝑘𝐢 minus six 𝐣 and that it moves from point 𝐴 with coordinates
negative two, negative two to point 𝐵 with coordinates six, 10.
The displacement 𝐝 is therefore
equal to the vector 𝐀𝐁. And we can calculate this by
subtracting the position vector of point 𝐴 from the position vector of point
𝐵. We have six 𝐢 plus 10𝐣 minus
negative two 𝐢 minus two 𝐣. And this is equal to eight 𝐢 plus
12𝐣. The work done will be equal to the
scalar product of the two vectors 𝑘𝐢 minus six 𝐣 and eight 𝐢 plus 12𝐣. However, at present, we don′t know
the value of the constant 𝑘.
We are told, however, that the
force 𝐅 acts in the opposite direction to the displacement 𝐀𝐁. This means that it must travel in
the same direction as the vector 𝐁𝐀, which is equal to negative eight 𝐢 minus
12𝐣. Comparing the 𝐢- and 𝐣-components
of this vector with vector 𝐅, we notice that the 𝐣-component has been multiplied
by the scalar two. In order for 𝐅 to have the same
direction as 𝐁𝐀, the same must be true of the 𝐢-components. 𝑘 multiplied by two or two 𝑘 must
be equal to negative eight.
Dividing through by two, we have 𝑘
is equal to negative four. And this means that the force
vector is equal to negative four 𝐢 minus six 𝐣. Clearing some space, we can now
calculate the work done. It is equal to the scalar product
of negative four 𝐢 minus six 𝐣 and eight 𝐢 plus 12𝐣. The scalar product is equal to the
sum of the products of the individual components, giving us negative four multiplied
by eight plus negative six multiplied by 12. This is equal to negative 32 minus
72, which gives us a final answer of negative 104.
Since there are no units given in
this question, we can conclude that the work done by the force 𝐅 is negative 104
units of work.
