A train engine of mass m starts moving such t | vedclass

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(A) According to the work-energy theorem,the total work done by all forces is equal to the change in kinetic energy: $W = \Delta KE = \frac{1}{2} m v^

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

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(A) According to the work-energy theorem,the total work done by all forces is equal to the change in kinetic energy: $W = \Delta KE = \frac{1}{2} m v^2$.Given $v = k\sqrt{s}$,we have $v^2 = k^2 s$.Differentiating with respect to time $t$: $2v \frac{dv}{dt} = k^2 \frac{ds}{dt}$.Since $\frac{ds}{dt} = v$,we get $2v \frac{dv}{dt} = k^2 v$,which simplifies to $\frac{dv}{dt} = \frac{k^2}{2}$.Integrating with respect to time $t$ (starting from rest at $t=0$): $v = \int_0^t \frac{k^2}{2} dt = \frac{k^2}{2} t$.Now,substituting $v$ into the work equation: $W = \frac{1}{2} m \left( \frac{k^2}{2} t \right)^2 = \frac{1}{2} m \left( \frac{k^4}{4} t^2 \right) = \frac{1}{8} m k^4 t^2$.

Column-$I$	Column-$II$
$(1)$ Potential Energy $=$ Kinetic Energy	$(a)$ At height $\frac{2h}{3}$
$(2)$ Potential Energy $= 2 \times$ Kinetic Energy	$(b)$ Constant at every point
	$(c)$ At height $\frac{h}{2}$

$A$ train engine of mass $m$ starts moving such that its velocity varies as $v = k\sqrt{s}$,where $k$ is a constant and $s$ is the distance covered. What is the total work done by all the forces acting on the engine in the first $t$ seconds?

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$(c)$ At height $\frac{h}{2}$

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