Question

Suppose a box of mass m is being pulled across the ground with a horizontal force fpull=ax2 , where a is a constant, from the position x=0 to x=d . find an expression for the work done by fpull on the box.

Answer 1

The work done by the force
`\(F_{\text{pull}} = a x^2\)` on the box moving from
`\(x = 0\)` to
`\(x = d\)` is
`\((1)/(3) a d^3\).`

Step-by-step explanation:

To calculate the work done by the force
`\(F_{\text{pull}}\)` as the box moves from
`\(x = 0\) to \(x = d\)`, we utilize the formula for work done by a variable force:
`\(W = \int F \cdot dx\)` . Here,
`\(F_{\text{pull}} = a x^2\)` is the force acting over the displacement
`\(dx\)` as the box moves from
`\(x = 0\) to \(x = d\)`. Integrating this force over the displacement gives us the work done.

The integral of
`\(a x^2\)` with respect to
`\(x\)` is
`\((a)/(3) x^3\)`. Evaluating this integral from
`\(x = 0\) to \(x = d\)` gives us the work done, which is
`\((a)/(3) d^3 - (a)/(3) \cdot 0^3\)`. Simplifying this expression results in
`\((1)/(3) a d^3\)` as the work done by the force
`\(F_{\text{pull}}\)` on the box from
`\(x = 0\) to \(x = d\)`.

Therefore, the work done by the force
`\(F_{\text{pull}} = a x^2\)` on the box as it moves from
`\(x = 0\) to \(x = d\) is \((1)/(3) a d^3\)`. This result indicates that the work done is directly proportional to the cube of the displacement
`\(d\)` and the constant
`\(a\)` determining the force's magnitude.