QJEEM23Phf561: JEE Main 2023 Physics - Class 11 JEEM23Phf561 Solution: A particle of mass moves in a straight line with retardation ,…

The work done against the retarding force is indeed equal to the loss in kinetic energy.

The force acting on the particle due to retardation is given by $F = ma = -2mx$ .

When we integrate this force over the displacement from $0$ to $x$ , we get:

$\Delta KE = W = \int F \cdot dx = \int (-2mx) \, dx = -mx^2$

The negative sign indicates that this is a loss of kinetic energy.

The problem states that the loss in kinetic energy is also given by $\left(\frac{10}{x}\right)^{-n}$ J. Therefore, we have:

$-mx^2 = \left(\frac{10}{x}\right)^{-n}$

Because this is a loss of kinetic energy, we should consider the absolute value. Hence,

$mx^2 = \left(\frac{10}{x}\right)^{-n}$

Substituting the given mass $m = 10 \, \text{g} = 0.01 \, \text{kg}$ , we get:

$0.01x^2 = \left(\frac{10}{x}\right)^{-n}$

This simplifies to:

$x^2 = \left(\frac{10}{x}\right)^{-n}$

Comparing the two sides, we can see that $n = 2$ .