Video Transcript
A variable force 𝐹, measured in newtons, is acting on a body, where 𝐹 is equal to three 𝑠 squared minus five. Find the work done by this force in the interval from 𝑠 equals four meters to 𝑠 equals five meters.
In this question, a variable force acts on an object. Since both the motion of the object and the force acting on it are in one dimension, we can use the formula 𝑊 is equal to the integral of 𝐹 with respect to 𝑠, where 𝑊 is the work done, 𝐹 is the force acting on the object, and 𝑠 is its displacement. In this question, we are told that 𝐹 is equal to three 𝑠 squared minus five. And we need to find the work done when it moves between 𝑠 equals four and 𝑠 equals five meters.
We can therefore use a definite integral with the values of four and five as the limits. We need to integrate three 𝑠 squared minus five with respect to 𝑠 between the limits of four and five. Using the power rule of integration, we can integrate term by term. Integrating three 𝑠 squared with respect to 𝑠 gives us three 𝑠 cubed over three, and this simplifies to 𝑠 cubed. Integrating the constant negative five with respect to 𝑠 gives us negative five 𝑠.
We now need to substitute the limits and find the difference between the two values. The work done is equal to five cubed minus five multiplied by five minus four cubed minus four multiplied by five. Five cubed is 125, and subtracting 25 from this gives us 100. Four cubed is 64, and subtracting 20 from this is 44. 𝑊 is therefore equal to 100 minus 44, which is equal to 56.
Since the force was measured in standard units of newtons and the displacement in meters, we can conclude that the work done by the force is equal to 56 joules.
