An object starts from rest and is acted upon | vedclass

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Question

(A) The $F-x$ graph is a straight line. The equation of the line is $F = x 	an \alpha - F_0$.According to Newton's second law,$F = ma = m v \frac{dv}

Vedclass · Accepted Answer

$\frac{2 F_0}{	an \alpha}$

Vedclass · Answer

(A) The $F-x$ graph is a straight line. The equation of the line is $F = x 	an \alpha - F_0$.According to Newton's second law,$F = ma = m v \frac{dv}{dx}$.Thus,$m v \frac{dv}{dx} = x 	an \alpha - F_0$.Integrating both sides with respect to $x$ from $x=0$ to $x=x_f$ (where velocity $v$ changes from $0$ to $0$):$\int_{0}^{0} m v \, dv = \int_{0}^{x_f} (x 	an \alpha - F_0) \, dx$.The left side is $0$,so:$0 = \left[ \frac{x^2}{2} 	an \alpha - F_0 x ight]_{0}^{x_f}$.$0 = \frac{x_f^2}{2} 	an \alpha - F_0 x_f$.Since $x_f 
eq 0$,we divide by $x_f$:$0 = \frac{x_f}{2} 	an \alpha - F_0$.$x_f = \frac{2 F_0}{	an \alpha}$.

An object starts from rest and is acted upon by a variable force $F$ as shown in the figure. If $-F_0$ is the initial value of the force at $x=0$,then the position of the object where it again comes to rest will be:

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