Answer:
a) The work required to stretch the spring from 20 centimeters to 25 centimeters is 0.313 joules, b) The area of the region enclosed by one loop of the curve [tex]r(\theta) = 2\cdot \sin 5\theta[/tex] is [tex]4\pi[/tex].
Explanation:
a) The work, measured in joules, is a physical variable represented by the following integral:
[tex]W = \int\limits^{x_{f}}_{x_{o}} {F(x)} \, dx[/tex]
Where
[tex]x_{o}[/tex], [tex]x_{f}[/tex] - Initial and final position, respectively, measured in meters.
[tex]F(x)[/tex] - Force as a function of position, measured in newtons.
Given that [tex]F = k\cdot x[/tex] and the fact that [tex]F = 25\,N[/tex] when [tex]x = 0.3\,m - 0.2\,m[/tex], the spring constant ([tex]k[/tex]), measured in newtons per meter, is:
[tex]k = \frac{F}{x}[/tex]
[tex]k = \frac{25\,N}{0.3\,m-0.2\,m}[/tex]
[tex]k = 250\,\frac{N}{m}[/tex]
Now, the work function is obtained:
[tex]W = \left(250\,\frac{N}{m} \right)\int\limits^{0.05\,m}_{0\,m} {x} \, dx[/tex]
[tex]W = \frac{1}{2}\cdot \left(250\,\frac{N}{m} \right)\cdot [(0.05\,m)^{2}-(0.00\,m)^{2}][/tex]
[tex]W = 0.313\,J[/tex]
The work required to stretch the spring from 20 centimeters to 25 centimeters is 0.313 joules.
b) Let be [tex]r(\theta) = 2\cdot \sin 5\theta[/tex]. The area of the region enclosed by one loop of the curve is given by the following integral:
[tex]A = \int\limits^{2\pi}_0 {[r(\theta)]^{2}} \, d\theta[/tex]
[tex]A = 4\int\limits^{2\pi}_{0} {\sin^{2}5\theta} \, d\theta[/tex]
By using trigonometrical identities, the integral is further simplified:
[tex]A = 4\int\limits^{2\pi}_{0} {\frac{1-\cos 10\theta}{2} } \, d\theta[/tex]
[tex]A = 2 \int\limits^{2\pi}_{0} {(1-\cos 10\theta)} \, d\theta[/tex]
[tex]A = 2\int\limits^{2\pi}_{0}\, d\theta - 2\int\limits^{2\pi}_{0} {\cos10\theta} \, d\theta[/tex]
[tex]A = 2\cdot (2\pi - 0) - \frac{1}{5}\cdot (\sin 20\pi-\sin 0)[/tex]
[tex]A = 4\pi[/tex]
The area of the region enclosed by one loop of the curve [tex]r(\theta) = 2\cdot \sin 5\theta[/tex] is [tex]4\pi[/tex].
