Question

mikaela is out for a bike ride on a breezy day. the wind blows out of the west such that it exerts a constant drag force (vec{f}_{drag}) of 115 n pointing east. initially riding north on flat roads, mikaela traverses a 1.30 km long circular arc at a constant speed that ends with her heading directly into the wind; the arc is a quarter circle that starts pointing north and ends pointing west, as shown in the figure. how much work (w) does the wind do on her as she rounds this curve from point a to point b?

Explanation:

Step1: Convert the distance unit

The length of the circular arc \(s = 1.30\space km=1.30\times10^{3}\space m\).

Step2: Use the work - formula \(W = F\cdot s\cdot\cos\theta\)

The force of wind \(F = 115\space N\), the angle \(\theta\) between the force of wind (east - direction) and the displacement (from \(A\) to \(B\), which has a component in the east - direction).
We can use the line - integral formula for work \(W=\int_{A}^{B}\vec{F}\cdot d\vec{r}\). Since \(\vec{F}\) is constant, \(W = F\int_{A}^{B}\cos\theta dr\).
For a circular arc, if we parameterize the displacement \(d\vec{r}\), and knowing that the \(x\) - component of the displacement (east - west direction) from \(A\) to \(B\) is \(r\) (the radius of the quarter - circle, and the length of the quarter - circle \(s=\frac{1}{2}\pi r\), but we can also use the fact that the dot - product \(\vec{F}\cdot d\vec{r}=Fdx\) (because \(\vec{F}\) is in the \(x\) (east) direction).
The radius of the quarter - circle \(r=\frac{s}{\frac{\pi}{2}}\), but another way: the work done by a constant force \(W = F\times x\) - component of the displacement.
The \(x\) - component of the displacement from \(A\) to \(B\) (for a quarter - circle) is \(r\), and \(s = \frac{\pi}{2}r\), \(r=\frac{2s}{\pi}\). But using the dot - product directly, if we consider the projection of the path on the direction of the force.
The work done by a constant force \(W=\vec{F}\cdot\Delta\vec{r}\). The force \(\vec{F}\) is in the east direction. The displacement has an east - ward component. For a quarter - circle, if we use the formula \(W = F\times r\) (where \(r\) is the radius of the quarter - circle and \(s=\frac{\pi}{2}r\), \(r = \frac{2s}{\pi}\)), but more simply, using the definition of work \(W=\int_{A}^{B}F\cos\theta ds\). Since \(\theta\) is the angle between \(\vec{F}\) and \(d\vec{s}\), and \(\cos\theta\) is the \(x\) (east) component of the unit vector along \(d\vec{s}\).
\(\int_{A}^{B}\cos\theta ds\) is the \(x\) - component of the displacement. For a quarter - circle starting from \(A\) (south - west, assume \(A\) is at \((r,0)\) in a coordinate system with \(x\) - axis east and \(y\) - axis north) and ending at \(B=(0,r)\), the \(x\) - component of the displacement is \(r\). And \(s=\frac{\pi}{2}r\), \(r=\frac{2s}{\pi}\). But another approach:
We know that \(W=\vec{F}\cdot\vec{d}\), where \(\vec{d}\) is the displacement vector. The force \(\vec{F}=(115,0,0)\space N\). The displacement vector from \(A\) to \(B\) (if \(A\) is at \((r,0)\) and \(B\) is at \((0,r)\)) has \(x\) - component \(-r\) (but wait, no, we should use the formula \(W=\int_{A}^{B}Fdx\).
Since \(F = 115\space N\) (constant) and the \(x\) - component of the displacement (from \(A\) to \(B\)): if we parameterize the path \(x = r\cos\varphi\), \(y = r\sin\varphi\) (\(\varphi\) from \(0\) to \(\frac{\pi}{2}\)), \(dx=-r\sin\varphi d\varphi\)
\(W=\int_{0}^{\frac{\pi}{2}}F(-r\sin\varphi)d\varphi\)
\(W=-Fr\int_{0}^{\frac{\pi}{2}}\sin\varphi d\varphi\)
\(W=-Fr[-\cos\varphi]_{0}^{\frac{\pi}{2}}\)
\(W=-Fr(- 0 + 1)\)
Since \(s=\frac{\pi}{2}r\), \(r=\frac{2s}{\pi}\)
\(W =-\frac{2Fs}{\pi}\)
Substitute \(F = 115\space N\) and \(s = 1.30\times 10^{3}\space m\)
\(W=-\frac{2\times115\times1.30\times 10^{3}}{\pi}\space J\approx - 9.51\times10^{4}\space J\)

Answer:

\(-9.51\times 10^{4}\)