### Video Transcript

A force 𝐅 equal to 𝑚𝐢 minus nine
𝐣 newtons acts on a particle, causing a displacement 𝐬 equal to negative five 𝐢
plus 𝑚 plus six 𝐣 centimetres. If the work done by the force is
0.02 joules, what is the value of 𝑚?

When the force and displacement are
given in vector form as in this case, we can calculate the work done by working out
the dot product of 𝐅 and 𝐬. The work done will be measured in
joules, the force needs to be measured in newtons, and the displacement in
metres. The force factor is in the correct
newtons, so 𝐅 is equal to 𝑚𝐢 minus nine 𝐣. Our displacement is given in
centimetres. To convert this to metres, we need
to divide each component by 100 as there are 100 centimetres in one metre. The displacement vector in metres
is therefore equal to negative five over 100 𝐢 plus 𝑚 plus six over 100 𝐣. This can be rewritten in decimal
form as negative 0.05𝐢 plus 0.01𝑚 plus 0.06 𝐣.

We can now calculate the work done
by finding the dot products of these two vectors. The dot product is found by firstly
multiplying the coefficients of 𝐢. 𝑚 multiplied by negative 0.05 is
equal to negative 0.05𝑚. We then multiply the coefficients
of 𝐣. Negative nine multiplied by 0.01𝑚
is negative 0.09𝑚. And negative nine multiplied by
0.06 is negative 0.54. We were told in the question that
the work done is 0.02. Therefore, this expression equals
0.02. Collecting like terms gives us
negative 0.14𝑚 minus 0.54 is equal to 0.02. At this stage, we might notice that
multiplying both sides by 100 will make the equation easier to deal with.

The equation simplifies to two is
equal to negative 14𝑚 minus 54. We can then add 54 to both sides so
that negative 14𝑚 is equal to 56. Finally, dividing both sides by
negative 14 will give us our value of 𝑚. 56 divided by 14 is equal to
four. Therefore, 56 divided by negative
14 is equal to negative four. The value of 𝑚 is negative
four. We could substitute this back in to
our initial expressions. The force is equal to negative four
𝐢 minus nine 𝐣. The displacement is equal to
negative five 𝐢 plus two 𝐣 centimetres or negative 0.05𝐢 plus 0.02𝐣 metres.