A bar of mass m resting on a smooth horizonta | vedclass

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Question

(A) The component of the force responsible for horizontal motion is $F \cos 	heta$.So,the acceleration of the bar is $a = \frac{F \cos 	heta}{m}$.Si

Vedclass · Accepted Answer

$v = \sqrt{\frac{2 F \sin 	heta}{m k}}$

Vedclass · Answer

(A) The component of the force responsible for horizontal motion is $F \cos 	heta$.So,the acceleration of the bar is $a = \frac{F \cos 	heta}{m}$.Since acceleration $a = v \frac{d v}{d x}$,we have:$v \frac{d v}{d x} = \frac{F \cos 	heta}{m}$Given $	heta = k x$,we have $d 	heta = k d x$,or $d x = \frac{d 	heta}{k}$.Substituting these into the equation:$v d v = \frac{F \cos 	heta}{m} \cdot \frac{d 	heta}{k}$Integrating both sides with initial conditions $v = 0$ at $	heta = 0$:$\int_{0}^{v} v d v = \frac{F}{m k} \int_{0}^{	heta} \cos 	heta d 	heta$$\frac{v^2}{2} = \frac{F}{m k} [\sin 	heta]_{0}^{	heta}$$\frac{v^2}{2} = \frac{F \sin 	heta}{m k}$$v^2 = \frac{2 F \sin 	heta}{m k}$$v = \sqrt{\frac{2 F \sin 	heta}{m k}}$

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