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Question
A particle of mass m moves on a straight line with its velocity varying with the distance travelled, according to the equation \[\nu = a\sqrt{x}\] , where a is a constant. Find the total work done by all the forces during a displacement from \[x = 0 \text{ to } x - d\] .
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Solution
Given,
\[\nu = a\sqrt{x} \left( \text{ uniformly accelerated motion } \right)\]
\[\text{ Displacement, s = d - 0 = d }\]
\[\text{ Putting x = 0, we get } \nu_1 = 0\]
\[\text{ Putting x = d, we get } \nu_2 = a\sqrt{d}\]
\[\alpha = \frac{\nu_2^2 - \nu_1^2}{2s} = \frac{a^2 d}{2d} = \frac{a^2}{2}\]
\[\text{ Force, F = m} \alpha = \frac{m a^2}{2}\]
\[\text{ Work done, W = Fs } \cos \theta\]
\[ = \frac{m a^2}{2} \times d = \frac{m a^2 d}{2}\]
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