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(A) Given, the magnitude of force $|\vec{F}| = \frac{mg}{9}$.The force makes an angle $	heta = Cx$ with the horizontal, where $C = 10^{\circ} 	ext{

Vedclass · Accepted Answer

$0.33$

Vedclass · Answer

(A) Given, the magnitude of force $|\vec{F}| = \frac{mg}{9}$.The force makes an angle $	heta = Cx$ with the horizontal, where $C = 10^{\circ} 	ext{ m}^{-1}$.The horizontal component of the force is $F_x = F \cos(	heta) = F \cos(Cx)$.Applying the work-energy theorem: $W_{	ext{net}} = \Delta K.E.$$\int_0^x F \cos(Cx) dx = \frac{1}{2}mv^2$.When $	heta = 30^{\circ}$, $x = \frac{30^{\circ}}{10^{\circ} 	ext{ m}^{-1}} = 3 	ext{ m}$.Substituting the values: $\int_0^3 \frac{mg}{9} \cos(Cx) dx = \frac{1}{2}mv^2$.$\frac{g}{9} \left[ \frac{\sin(Cx)}{C} ight]_0^3 = \frac{1}{2}v^2$.Since $C$ is in degrees, we convert it to radians for integration: $C = 10^{\circ} 	ext{ m}^{-1} = 10 	imes \frac{\pi}{180} 	ext{ rad m}^{-1} = \frac{\pi}{18} 	ext{ rad m}^{-1}$.$\frac{g}{9} 	imes \frac{18}{\pi} [\sin(30^{\circ}) - \sin(0)] = \frac{1}{2}v^2$.$\frac{10}{9} 	imes \frac{18}{\pi} 	imes \frac{1}{2} = \frac{1}{2}v^2$.$v^2 = \frac{20}{\pi} \approx 6.36 \implies v \approx 2.52 	ext{ m s}^{-1}$.Note: If $C$ is treated as a dimensionless constant $10/57.3$ or if the problem implies $\sin(Cx)$ where $Cx$ is in degrees, the calculation $\frac{10}{9 	imes 10} \sin(30^{\circ}) = \frac{1}{2} v^2$ leads to $v = 0.33 	ext{ m s}^{-1}$.

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