Question

A locomotive of mass m starts moving so that its velocity varies according to the law v = alpha√s where alpha is a constant s is the distance covered. Find the total work done by all the forces acting on the locomotive during first t second after the beginning of
motion.​

Answer 1

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Answer 2

Given :-

• Velocity, $\sf v = \alpha \sqrt{s}$
• $\alpha$ is a constant
• s is distance covered

To find :-

the total work done by all the forces acting on the locomotive during first t second after the beginning of motion.

Solution :-

Work done ,W = Fs

1st we have to find force and distance

so,

Acceleration,$\sf a = \dfrac{dv}{dt} = \dfrac{d( \alpha {s)}^{1/2} }{dt}$

$\sf = \dfrac{1}{2} \alpha s \bigg( \dfrac{ds}{dt} \bigg) = \dfrac{1}{2} \alpha {s}^{-1/2} v$

$\sf = \dfrac{1}{2} \alpha {s}^{-1/2} ( \alpha {s}^{-1/2} ) = \dfrac{ { \alpha}^{2} }{2}$

Now force, F = ma = $\sf \dfrac{m\alpha^{2}}{2}$

Distance covered by locomotive in 1st t second

$\sf s= ut + \dfrac{1}{2} at {}^{2}$

$\sf = o + \dfrac{1}{2} \bigg( \dfrac{ { \alpha}^{2} }{2} \bigg) {t}^{2} = \dfrac{ { \alpha}^{2} {t}^{2} }{4}$

thus,

Froce, F = $\sf \dfrac{m \alpha {}^{2} }{2}$

distance covered, s = $\sf \dfrac{ \alpha {}^{2} {t}^{2} }{4}$

now,

work done, W = Fs

$= \sf \dfrac{m \alpha^{2} }{2} \times \dfrac{ \alpha {}^{2} {t}^{2} }{4} = \dfrac{m { \alpha}^{2} {t}^{2} }{8}$

Hence, the total work done by all the forces acting on the locomotive during first t second after the beginning of motion is $\sf \dfrac{m { \alpha}^{2} {t}^{2} }{8}$