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Physics Question on work, energy and power

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

Answer

 Given, a=−2x ⇒vdvdx=−2x ⇒vdv=−2xdx ⇒∫v1v2vdv=−2∫0xxdx ⇒v222−v122=−2x22 ⇒mv122−mv222=mx2=101000x2=10−2x2=(10x)−2 n=2.\begin{aligned} & \text { Given, } a=-2 x \\\ & \Rightarrow \frac{v d v}{d x}=-2 x \\\ & \Rightarrow v d v=-2 x d x \\\ & \Rightarrow \int_{v_1}^{v_2} v d v=-2 \int_0^x x d x \\\ & \Rightarrow \frac{v_2^2}{2}-\frac{v_1^2}{2}=-\frac{2 x^2}{2} \\\ & \Rightarrow \frac{m v_1^2}{2}-\frac{m v_2^2}{2}=m x^2=\frac{10}{1000} x^2=10^{-2} x^2=\left(\frac{10}{x}\right)^{-2} \\\ & \mathrm{n}=2 . \end{aligned}