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(B) According to Newton's second law of motion,the horizontal component of the force causes acceleration $a_{acc}$ along the surface:$F \cos 	heta =

Vedclass · Accepted Answer

$v = 2(\sin 	heta - \sin a)^{1/2}$

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(B) According to Newton's second law of motion,the horizontal component of the force causes acceleration $a_{acc}$ along the surface:$F \cos 	heta = m \frac{dv}{dt}$Given $|F| = 2mb$ and $	heta = a + bs$,we have:$2mb \cos 	heta = m \frac{dv}{dt} \implies \frac{dv}{dt} = 2b \cos 	heta$Using the chain rule $\frac{dv}{dt} = v \frac{dv}{ds}$:$v \frac{dv}{ds} = 2b \cos 	heta$Since $	heta = a + bs$,we have $d	heta = b ds$,or $ds = \frac{d	heta}{b}$. Substituting this:$v dv = 2b \cos 	heta \left( \frac{d	heta}{b} ight) = 2 \cos 	heta d	heta$Integrating both sides from the initial state ($v=0$ at $	heta=a$) to the final state ($v$ at $	heta$):$\int_{0}^{v} v dv = \int_{a}^{	heta} 2 \cos 	heta d	heta$$\left[ \frac{v^2}{2} ight]_{0}^{v} = 2 [\sin 	heta]_{a}^{	heta}$$\frac{v^2}{2} = 2(\sin 	heta - \sin a)$$v^2 = 4(\sin 	heta - \sin a)$$v = 2(\sin 	heta - \sin a)^{1/2}$

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