A can of sardines is made to move along an x axis from x = 0.25 m to x = 1.25 m by a force with a

A can of sardines is made to move along an x axis from x = 0.25 m to x = 1.25 m by a force with a magnitude given by F = exp(-4x2), with x in meters and F in newtons. (Here exp is the exponential function.) How much work is done on the can by the force?

"To calculate the work done on the can by the force given by \( F = \exp(-4x^2) \), we need to evaluate the work integral from the initial position \( x_1 = 0.25 \, \text{m} \) to the final position \( x_2 = 1.25 \, \text{m} \).

Work Done

The work done by a variable force is given by the integral:
\[
W = \int_{x_1}^{x_2} F(x) \, dx = \int_{0.25}^{1.25} \exp(-4x^2) \, dx
\]

Unfortunately, this integral involves a non-elementary function, but it is related to the error function, \(\text{erf}(x)\). The solution must be expressed in terms of this function or approximated numerically.

Calculation via Error Function

The integral of the given force can be expressed using the error function \(\text{erf}(x)\):
\[
\int \exp(-ax^2) \, dx = \frac{\sqrt{\pi}}{2\sqrt{a}} \text{erf}(\sqrt{a}x) + C
\]

For our specific integral \(\exp(-4x^2)\), \(a = 4\):
\[
\int \exp(-4x^2) \, dx = \frac{\sqrt{\pi}}{4} \text{erf}(2x) + C
\]

Evaluating from \(x = 0.25\) to \(x = 1.25\):
\[
W = \frac{\sqrt{\pi}}{4} \left[ \text{erf}(2 \times 1.25) - \text{erf}(2 \times 0.25) \right]
\]

Approximation Using Values

Using standard values for the error function:
\(\text{erf}(2.5) \approx 0.999593\)
\(\text{erf}(0.5) \approx 0.520500\)

Substitute to find the work:
\[
W = \frac{\sqrt{\pi}}{4} \left( 0.999593 - 0.520500 \right)
\]
\[
W = \frac{\sqrt{\pi}}{4} \times 0.479093 \approx 0.2110 \, \text{J}
\]

The work done on the can by the force is approximately \(0.2110 \, \text{J}\)."