A particle of mass $10 \mathrm{~g}$ 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 $\left(\frac{10}{x}\right)^{-n}$ J. The value of $\mathrm{n}$ will be __________

<p>The work done against the retarding force is indeed equal to the loss in kinetic energy. </p> <p>The force acting on the particle due to retardation is given by $F = ma = -2mx$. </p> <p>When we integrate this force over the displacement from $0$ to $x$, we get:</p> <p>$\Delta KE = W = \int F \cdot dx = \int (-2mx) \, dx = -mx^2$</p> <p>The negative sign indicates that this is a loss of kinetic energy. </p> <p>The problem states that the loss in kinetic energy is also given by $\left(\frac{10}{x}\right)^{-n}$ J. Therefore, we have:</p> <p>$-mx^2 = \left(\frac{10}{x}\right)^{-n}$</p> <p>Because this is a loss of kinetic energy, we should consider the absolute value. Hence,</p> <p>$mx^2 = \left(\frac{10}{x}\right)^{-n}$</p> <p>Substituting the given mass $m = 10 \, \text{g} = 0.01 \, \text{kg}$, we get:</p> <p>$0.01x^2 = \left(\frac{10}{x}\right)^{-n}$</p> <p>This simplifies to:</p> <p>$x^2 = \left(\frac{10}{x}\right)^{-n}$</p> <p>Comparing the two sides, we can see that $n = 2$. </p>