Complete the following. If a force given by 𝐹 of 𝑥 acts in the direction of the positive 𝑥-axis, then the work done by the force in moving a particle from 𝑥 equals 𝑎 to 𝑥 equals 𝑏, where 𝑎 is less than 𝑏, equals what.
We begin by recalling that for a constant force vector 𝐅 that acts on an object as that object undergoes a displacement vector 𝐬, then the work done by the force 𝑊 is the scalar product of the force and displacement. This can also be written as 𝑊 is equal to 𝐹 multiplied by 𝑠 multiplied by cos 𝜃, where 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.
If, on the other hand, the force varies as the object moves, in this case as the function 𝐹 of 𝑥, then the graph of 𝐹 cos 𝜃 against 𝑠 might look something like the figure shown. As the work done by the force will be equal to the area under the curve, we can calculate this by integrating 𝐹 cos 𝜃 with respect to 𝑠.
In this question, we are told that the force acts in the direction of the positive 𝑥-axis. This means that the force and displacement are in the same direction. And 𝜃 is therefore equal to zero degrees. The cos of zero degrees is equal to one. And this means that the work done 𝑊 is equal to the integral of the force 𝐹 of 𝑥 with respect to 𝑥.
We need to find the work done in moving the particle from 𝑥 equals 𝑎 to 𝑥 equals 𝑏. And since 𝑎 is less than 𝑏, it will appear on the left of 𝑏 on the positive 𝑥-axis. This means that we will have a definite integral with lower limit equal to 𝑎 and upper limit equal to 𝑏. The work done by the force 𝐹 of 𝑥 in moving the particle from 𝑥 equals 𝑎 to 𝑥 equals 𝑏 can be calculated by working out the shaded area in the figure. And this is equal to the integral of 𝐹 of 𝑥 with respect to 𝑥 between the limits 𝑎 and 𝑏.