A Beautiful Problem

Classical Mechanics Level 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 which are acting on the locomotive during the first $t$ seconds, after the beginning of motion.

If $W=\large\frac { m{ k }^{ \alpha }{ t }^{ \beta } }{ \mu }$ , find $\alpha +\beta +\mu$ .

The answer is 14.

1 solution

Rishabh Jain
Jul 6, 2016

Since $v=\dfrac{ds}{dt}$ ,

$\dfrac{ds}{dt}=k\sqrt s\implies \int_0^s \dfrac{ds}{\sqrt s}=k\int_0^t dt$

Solving we get: $\sqrt s=\dfrac{kt}{2}\text{ OR } s=\dfrac{k^2t^2}4$

$\therefore v=\dfrac{ds}{dt}=\dfrac{k^2t}{2}$

By Work-K•E change theorem:

$W=\Delta K=\dfrac{m}2\left(\left(\dfrac{k^2t}{2}\right)^2-0\right)=\dfrac{mk^4t^2}{8}$ .

Hence, $\large \therefore 4+2+8=\boxed{\color{#3D99F6}{14}}$

An alternate and slightly easier solution would be -

$W = \int_{0}^{s} F ds = \int_{0}^{s} ma \cdot ds$

Also, $v^{2} = k^{2}s = 2as \Rightarrow a = \frac {k^{2}}{2}$ . Since $s = \frac {1}{2}at^{2} = \frac {k^{2}t^{2}}{4}$

$W = \frac {mk^{2}}{2}\int_{0}^{s} ds = \frac {mk^{2}s}{2} = \boxed {\frac {mk^{4}t^{2}}{8}}$

N. Aadhaar Murty - 8 months, 3 weeks ago

I was confused when I was working on the problem. I thought the character in the denominator was intended to be the coefficient of friction. I got the value of 8, just as you did, but thought I was wrong because coefficient of friction wouldn’t be 8.

Matt Owings - 5 months, 3 weeks ago

A Beautiful Problem

The answer is 14.

1 solution

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