True or False: If an object moves
along a straight line by a varying force 𝑓 of 𝑥, then the work done 𝑊 to move the
object from 𝑥 equals 𝑎 to 𝑥 equals 𝑏 is given by 𝑊 is the integral of 𝑓 of 𝑥
with respect to 𝑥 from 𝑏 to 𝑎.
We begin by recalling that if the
force acting on an object is described by a continuous function, such as in the
graph shown, we have to use integration to find the area under the curve and hence
the work done. In the graph drawn, 𝐹 is the
magnitude of the force, 𝑆 is the magnitude of the displacement, and 𝜃 is the angle
between the force acting on the object and its displacement.
In this question, we are told that
the object moves along a straight line by a varying force. And we know that if the force and
displacement are in the same direction, then 𝜃 is equal to zero degrees. The cos of zero degrees is one. We can then relabel our axes as the
displacement 𝑥 and the varying force 𝑓 of 𝑥. To calculate the work done or the
area under the curve, we will need to integrate the function 𝑓 of 𝑥 with respect
to 𝑥. This suggests that the statement
might be true.
However, we are interested in the
work done to move the object from 𝑥 equals 𝑎 to 𝑥 equals 𝑏. And in order to calculate the
shaded area, we will have a definite integral with lower limit 𝑎 and upper limit
𝑏. The limits in the expression in the
question are the opposite way round. This expression would give us the
work done to move the object from 𝑥 equals 𝑏 to 𝑥 equals 𝑎. And we can therefore conclude that
the statement is false.