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Q.

A particle of mass 10g moves in a straight line with retardation $2 x$ , where $x$ is the displacement in SI units. Its loss of kinetic energy for above displacement is ${(\frac{10}{x})}^{- n} J$ . The value of $n$ will be____

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answer is 2.

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Detailed Solution

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$m = 10 g a = - 2 x V d V = - 2 x d x$ integrating both side

$\frac{V^{2}}{2} = - x^{2}, {(V^{2})}_{V_{1}}^{V_{2}} = {(- 2 x^{2})}_{0}^{x}; V_{2}^{2} - V_{1}^{2} = - 2 x^{2}$

$Loss in kinetic energy = \frac{1}{2} m (V_{1}^{2} - V_{2}^{2})$

$\begin{array}{l} = \frac{1}{2} m (2 x^{2}) = m x^{2} \\ = 10^{- 2} x^{2} = {(\frac{10}{x})}^{- 2} \\ n = 2 \end{array}$

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