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Problem 46 Hard Difficulty

Work done by a radial force with constant magnitude A particle moves along the smooth curve $y=f(x)$ from $(a, f(a))$ to $(b, f(b)) .$ The force moving the particle has constant magnitude $k$ and always points away from the origin. Show that the work done by the force is $$\int_{C} \mathbf{F} \cdot \mathbf{T} d s=k\left[\left(b^{2}+(f(b))^{2}\right)^{1 / 2}-\left(a^{2}+(f(a))^{2}\right)^{1 / 2}\right]$$

Answer

$$
\begin{aligned} \text { work } &=\int \mathbf{F} . \mathbf{T} d s \\ &=\int_{C} \mathbf{F} \cdot \frac{d \mathbf{r}}{d x} d x \end{aligned}
$$
$$
\begin{array}{c}{=k \int_{a}^{b} \frac{d}{d x}\left(\sqrt{x^{2}+[f(x)]^{2}}\right) d x} \\ {=k\left(\sqrt{x^{2}+[f(x)]^{2}}\right)_{a}^{b}}\end{array}
$$
$$
\begin{array}{l}{=k\left(\sqrt{b^{2}+[f(b)]^{2}}-\sqrt{a^{2}+[f(a)]^{2}}\right)} \\ {=k\left[\left(b^{2}+(f(b))^{2}\right)^{\frac{1}{2}}-\left(a^{2}+(f(a))^{2}\right)^{2}\right]} \\ {\text { Hence, proved. }}\end{array}
$$

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Video Transcript

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