A body of mass m = 10^{-2} kg is moving in a | vedclass

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(C) Given,the final energy is $\frac{1}{2} m v_f^2 = \frac{1}{8} m v_0^2$. This implies $v_f^2 = \frac{1}{4} v_0^2$,so $v_f = \frac{v_0}{2}$. From New

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$10^{-4} \ kg \ m^{-1}$

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(C) Given,the final energy is $\frac{1}{2} m v_f^2 = \frac{1}{8} m v_0^2$. This implies $v_f^2 = \frac{1}{4} v_0^2$,so $v_f = \frac{v_0}{2}$. From Newton's second law,$m \frac{dv}{dt} = -kv^2$. Rearranging the terms,we get $\frac{dv}{v^2} = -\frac{k}{m} dt$. Integrating both sides: $\int_{v_0}^{v_0/2} v^{-2} dv = -\frac{k}{m} \int_{0}^{10} dt$. Evaluating the integrals: $\left[ -\frac{1}{v} ight]_{v_0}^{v_0/2} = -\frac{k}{m} [t]_0^{10}$. Substituting the limits: $-\left( \frac{2}{v_0} - \frac{1}{v_0} ight) = -\frac{k}{m} (10)$. This simplifies to $\frac{1}{v_0} = \frac{10k}{m}$. Solving for $k$: $k = \frac{m}{10v_0} = \frac{10^{-2}}{10 	imes 10} = 10^{-4} \ kg \ m^{-1}$.

$A$ body of mass $m = 10^{-2} \ kg$ is moving in a medium and experiences a frictional force $F = -kv^2$. Its initial speed is $v_0 = 10 \ ms^{-1}$. If,after $10 \ s$,its energy is $\frac{1}{8} mv_0^2$,the value of $k$ will be

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