A block of mass 1 kg,moving along the x-axis | vedclass

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(D) Given: mass $m = 1 \ kg$,initial velocity $v_{i} = 10 \ m/s$,force $F = -kx$,where $k = 10 \ N/m$. Using Newton's second law,$a = \frac{F}{m} = -\

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$8$

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(D) Given: mass $m = 1 \ kg$,initial velocity $v_{i} = 10 \ m/s$,force $F = -kx$,where $k = 10 \ N/m$. Using Newton's second law,$a = \frac{F}{m} = -\frac{kx}{m} = -\frac{10x}{1} = -10x$. We know that $a = v \frac{dv}{dx}$,so $v \frac{dv}{dx} = -10x$. Integrating both sides: $\int_{10}^{v} v \, dv = \int_{0.1}^{1.9} -10x \, dx$. $\left[ \frac{v^2}{2} \right]_{10}^{v} = -10 \left[ \frac{x^2}{2} \right]_{0.1}^{1.9}$. $\frac{v^2 - 100}{2} = -5 \left( 1.9^2 - 0.1^2 \right)$. $\frac{v^2 - 100}{2} = -5 \left( 3.61 - 0.01 \right) = -5 \left( 3.60 \right) = -18$. $v^2 - 100 = -36$. $v^2 = 64$. $v = 8 \ m/s$.

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