Question
A locomotive of mass $m$ starts moving so that its velocity varies as $v=\alpha s^{2 / 3}$, where $\alpha$ is a constant and $s$ is the distance traversed. The total work done by all the forces acting on the locomotive during the first $t$ second after the start of motion is(A) $\frac{1}{8} m \alpha^{4} t^{2}$(B) $\frac{m \alpha^{6} t^{4}}{162}$(C) $\frac{m \alpha^{6} t^{4}}{81}$(D) $\frac{m \alpha^{4} t^{2}}{2}$
Step 1
We also know that velocity is the rate of change of displacement, i.e., $v = \frac{ds}{dt}$. So, we can write the given equation as $\frac{ds}{dt} = \alpha s^{2 / 3}$. Show more…
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Differentating $v(s)$ with respect to time $$ \frac{d v}{d t}=\frac{a}{2 \sqrt{s}} \frac{d s}{d t}=\frac{a}{2 \sqrt{s}} a \sqrt{s}=\frac{a^{2}}{2}=w $$ (As locomotive is in unidrectional motion) Hence force acting on the locomotive $F=m w=\frac{m a^{2}}{2}$ Let, at $v=0$ at $t=0$ then the distance covered during the first $t$ seconds $$ s=\frac{1}{2} w t^{2}=\frac{1}{2} \frac{a^{2}}{2} t^{2}=\frac{a^{2}}{4} t^{2} $$ Hence the sought work, $A=F s=\frac{m a^{2}}{2} \frac{\left(a^{2} t^{2}\right)}{4}=\frac{m a^{4} t^{2}}{8}$
Physical Fundamentals Of Mdchanics
Laws of Conservation of Energy, Momemtum, and Angular Momentum
$\Lambda$ 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 $f-0, s-0 \quad \therefore \quad c-0$ $\therefore t-\frac{2}{a} \sqrt{s}$ $$ \text { or } t^{2}-\frac{4 s}{a^{2}} $$ $\therefore W-\frac{1}{2} m a^{2} \frac{a^{2} t^{2}}{4}-\frac{1}{8} m a^{4} t^{2}$
Work, Energy, Power and Circular Motion
Section A
$\Lambda$ train accelerates from rest for tine $t_{1}$ at a constant rate $\alpha$ for distance $x$. Then, it decelerates to rest at constant rate $\beta$ in time $t_{2}$ for distance $y$. I'hen, (a) $\frac{x}{y}-\frac{\beta}{\alpha}$ (b) $\frac{\beta}{\alpha}-\frac{t_{1}}{t_{2}}$ (c) $\frac{x}{y}-\frac{t_{1}}{t_{2}}$ (d) $x-y$
Motions in One Dimension
Section B
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