Video Transcript
A particle moved from point 𝐴
seven, negative three to point 𝐵 negative nine, two along a straight line under the
action of a force 𝐅 of magnitude eight root 10 newtons acting in the same direction
as the vector 𝐜 equals negative three 𝐢 minus 𝐣. Calculate the work done by the
force, given that the magnitude of the displacement is measured in meters.
Remember, the work done by a force
of vector 𝐅 over some displacement vector 𝐝 is the dot product or scalar product
of those two vectors. In this question, we’re told that
the particle moves from point 𝐴 to point 𝐵, so we can calculate the displacement
of the particle fairly easily. In vector form, we’ll say that the
displacement 𝐝 is the vector 𝐀𝐁, and that’s the vector 𝐎𝐁 minus the vector
𝐎𝐀. Now, since we know the position
from the origin, that’s the point zero, zero, then the vector 𝐎𝐁 is simply
negative nine 𝐢 plus two 𝐣 and the vector 𝐎𝐀 is seven 𝐢 minus three 𝐣. So, the vector 𝐝 is the difference
of these, and we can find that by subtracting the individual components. Negative nine minus seven is
negative 16, and two minus negative three is five 𝐣. So, we have the displacement
vector. It’s negative 16𝐢 plus five
𝐣.
How do we calculate the force
though? We know it has a magnitude of eight
root 10, but we don’t know its vector. We do, however, know it acts in the
same direction as this vector here, negative three 𝐢 minus 𝐣. So, let’s sketch this out. Vector 𝐜 looks a little something
like this. For every three units left, we move
one unit down. Vector 𝐅 acts in the same
direction as this but has a magnitude of eight root 10. Once again, for every three units
left, we still move one unit down. So, we can say that if we define 𝐅
sub 𝑖 as being the 𝐢-component of our force and 𝐅 sub 𝑗 as being the
𝐣-component, then 𝐅 sub 𝑖 is equal to three times 𝐅 sub 𝑗.
And, of course, since the
𝐢-components and 𝐣-components of the vectors are perpendicular to one another, we
can form a right triangle and apply the Pythagorean theorem to find the value of 𝐅
sub 𝑖 and 𝐅 sub 𝑗. That is, the magnitude of 𝐅 sub 𝑖
squared plus the magnitude of 𝐅 sub 𝑗 squared is equal to eight root 10 all
squared. Now, of course, 𝐅 sub 𝑖 is three
times 𝐅 sub 𝑗. So, we can replace the magnitude of
𝐅 sub 𝑖 with three times the magnitude of 𝐅 sub 𝑗. Similarly, eight root 10 all
squared is 640. Three squared is nine, so we have a
total of 10 lots of 𝐅 sub 𝑗 squared. Then, we divide through by 10. And finally, we’ll take the square
root of both sides. Now, we only need to take the
positive square root of 64 because we’re looking at the magnitude, which is simply a
length. And so, the magnitude of 𝐅 sub 𝑗
is equal to eight. Since the magnitude of 𝐅 sub 𝑖 is
three times the magnitude of 𝐅 sub 𝑗, it’s three times eight, which is equal to
24. We know the direction in which our
force is acting. So we can define the vector force
to be negative 24𝐢 minus eight 𝐣 newtons.
Let’s clear some space and we’re
ready to take the dot product of our force and our displacement. The work done is the dot or scalar
product of negative 24𝐢 minus eight 𝐣 and negative 16𝐢 plus five 𝐣. That’s negative 24 times negative
16 plus negative eight times five, and that’s equal to 344. The work done is in joules. So, we find the work done by our
force is 344 joules.
