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Question

(A) Given the velocity law $v = k\sqrt{S}$.We know that acceleration $a = v \frac{dv}{dS}$.Differentiating $v$ with respect to $S$: $\frac{dv}{dS} = k

Vedclass · Accepted Answer

$W = \frac{1}{8} mk^4 t^2$

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(A) Given the velocity law $v = k\sqrt{S}$.We know that acceleration $a = v \frac{dv}{dS}$.Differentiating $v$ with respect to $S$: $\frac{dv}{dS} = k \cdot \frac{1}{2\sqrt{S}} = \frac{k^2}{2v}$.Thus,$a = v \cdot \frac{k^2}{2v} = \frac{k^2}{2}$.Since the acceleration is constant,we can use the equation of motion $v = u + at$. Starting from rest $(u=0)$,$v = \frac{k^2 t}{2}$.According to the Work-Energy Theorem,the total work done by all forces is equal to the change in kinetic energy: $W = \Delta K.E. = \frac{1}{2}mv^2 - 0$.Substituting the value of $v$: $W = \frac{1}{2}m \left(\frac{k^2 t}{2}\right)^2 = \frac{1}{2}m \left(\frac{k^4 t^2}{4}\right) = \frac{1}{8}mk^4 t^2$.

$A$ locomotive of mass $m$ starts moving so that its velocity varies according to the law $v = k \sqrt{S}$,where $k$ is a constant and $S$ is the distance covered. Find the total work performed by all the forces acting on the locomotive during the first $t$ seconds after the beginning of motion.

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