Video Transcript
A body moves along the 𝑥-axis under the action of a force 𝐹. Given that 𝐹 is equal to 𝑥 cubed over 10 newtons, where 𝑥 meters is the displacement from the origin, determine the work done on the body by 𝐹 when the body moves from 𝑥 equals one meter to 𝑥 equals two meters.
We begin by recalling that the work done by a force on an object as the object moves along the path parallel to the force is given by 𝑊 is equal to the integral of 𝐹 with respect to 𝑥, where 𝑊 is the work done and 𝐹 is the magnitude of the force. In this question, we are told this is equal to 𝑥 cubed over 10 newtons.
We are asked to find the work done when it moves between 𝑥 equals one meter and 𝑥 equals two meters. This means that we will have a definite integral where these values are the limits. The work done is equal to the integral of 𝑥 cubed over 10 with respect to 𝑥 between the limits one and two. We know that integrating 𝑥 to the power of 𝑛 with respect to 𝑥 gives us 𝑥 to the power of 𝑛 plus one divided by 𝑛 plus one plus a constant 𝐶.
Integrating 𝑥 cubed over 10 gives us 𝑥 to the fourth power over four multiplied by 10. This is equal to 𝑥 to the fourth power over 40. And as we are dealing with a definite integral, there will be no constant of integration. Next, we substitute in the upper and lower limits. Subtracting our two answers, we have 16 over 40 minus one over 40, which is equal to 15 over 40.
Both the numerator and denominator are divisible by five. The work done on the body by 𝐹 when the body moves from 𝑥 equals one meter to 𝑥 equals two meters is therefore equal to three over eight or three-eighths joules.
