Λlocomotive of mass m staris moving so that its velocily...

$\Lambda$ locomotive of mass $m$ staris moving so that its velocily varics according to the law $v=a \sqrt{s}$, where $a$ is a constant, and $s$ is the distance covered. Find the total work performed by all the forees which are acting on the locomotive during the first $t$ scconds. Solution By the work-energy theorem $W$ (work done) $=$ change in the kinetic energy $$ -\frac{1}{2} m v^{2}-0-\frac{1}{2} m v^{2}-\frac{1}{2} m(a \sqrt{s})^{2}-\frac{1}{2} a^{2} s $$ $\because v-a \sqrt{s}, \frac{d s}{d t}-a \sqrt{s}\left(\because v-\frac{d s}{d t}\right) \Rightarrow d t-\frac{d s}{a \sqrt{s}} \Rightarrow \int d t-\int \frac{d s}{a \sqrt{s}}$ $\therefore t-\frac{2}{a} \sqrt{s} \mathrm{l} c$ when $f-0, s-0 \quad \therefore \quad c-0$ $\therefore t-\frac{2}{a} \sqrt{s}$ $$ \text { or } t^{2}-\frac{4 s}{a^{2}} $$ $\therefore W-\frac{1}{2} m a^{2} \frac{a^{2} t^{2}}{4}-\frac{1}{8} m a^{4} t^{2}$

$\Lambda$ locomotive of mass $m$ staris moving so that its velocily varics according to the law $v=a \sqrt{s}$, where $a$ is a constant, and $s$ is the distance covered. Find the total work performed by all the forees which are acting on the locomotive during the first $t$ scconds.
Solution By the work-energy theorem $W$ (work done) $=$ change in the kinetic energy
$$
-\frac{1}{2} m v^{2}-0-\frac{1}{2} m v^{2}-\frac{1}{2} m(a \sqrt{s})^{2}-\frac{1}{2} a^{2} s
$$
$\because v-a \sqrt{s}, \frac{d s}{d t}-a \sqrt{s}\left(\because v-\frac{d s}{d t}\right) \Rightarrow d t-\frac{d s}{a \sqrt{s}} \Rightarrow \int d t-\int \frac{d s}{a \sqrt{s}}$
$\therefore t-\frac{2}{a} \sqrt{s} \mathrm{l} c$
when $f-0, s-0 \quad \therefore \quad c-0$
$\therefore t-\frac{2}{a} \sqrt{s}$
$$
\text { or } t^{2}-\frac{4 s}{a^{2}}
$$
$\therefore W-\frac{1}{2} m a^{2} \frac{a^{2} t^{2}}{4}-\frac{1}{8} m a^{4} t^{2}$

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Key Concepts

Work-Energy Theorem

This principle states that the net work done on an object is equal to the change in its kinetic energy. It provides a way to compute work without explicitly dealing with the force applied at each point along an object's path, which is particularly useful when dealing with variable forces or speeds.

Kinetic Energy

Kinetic energy is the energy possessed by an object due to its motion, typically expressed as one half the product of its mass and the square of its velocity (½mv²). Understanding kinetic energy is essential since its change is directly tied to the work done on the object.

Variable Force and Non-Constant Acceleration

In problems where the force or acceleration varies with time, displacement, or another variable, the direct application of Newton’s second law becomes more complex. Instead, one must often integrate the force over the path, or relate variables through differential equations, to determine quantities like work.

Differential Equations in Kinematics

When motion is defined by non-linear relationships—such as a velocity that is a function of displacement—differential equations and their integration become necessary. These mathematical tools help establish relationships among displacement, velocity, and time, and are critical for solving for work done when acceleration is not constant.

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