Video Transcript
A body is moving in a straight line
from point 𝐴 negative six, zero to point 𝐵 negative five, four under the action of
the vector force 𝐅, which is equal to 𝑚𝐢 plus two 𝐣 newtons. Given that the change in the body’s
potential energy is two joules and that the displacement is in meters, determine the
value of the constant 𝑚.
We are told that the body moves in
a straight line from point 𝐴 to point 𝐵, where 𝐴 and 𝐵 have coordinates negative
six, zero and negative five, four. This means that we move one unit to
the right and four units up. If we consider the unit vectors 𝐢
and 𝐣 in the horizontal and vertical direction, respectively, our displacement
vector is equal to 𝐢 plus four 𝐣.
We are also told that the vector
force acting on the body is 𝑚𝐢 plus two 𝐣 newtons. We know that the work done is the
dot or scalar product of the force vector and the displacement vector. The work done is therefore equal to
the dot product of 𝐢 plus four 𝐣 and 𝑚𝐢 plus two 𝐣.
To calculate the dot product, we
find the sum of the products of the individual components. In this question, this is equal to
one multiplied by 𝑚 plus four multiplied by two. The 𝐢-components are one and 𝑚,
and the 𝐣-components are four and two. This simplifies to 𝑚 plus
eight.
We are also told in the question
that the change in potential energy is equal to two joules. As energy can only be transferred
and not destroyed or created, we know that the sum of the work done and the
gravitational potential energy is equal to zero. This means that 𝑚 plus eight plus
two must equal zero. Collecting like terms, we have 𝑚
plus 10 is equal to zero. Finally, we can subtract 10 from
both sides of this equation, giving us a value of 𝑚 equal to negative 10. This means that the vector force 𝐅
is equal to negative 10𝐢 plus two 𝐣.
