PYQ NEET- Work Energy And Power L-9

Question:#

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

Answer:#

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 $F=m a=-2 m x$. When we integrate this force over the displacement from 0 to $x$, we get: $$ \Delta KE=W=\int F \cdot dx=\int(-2 m x) dx=-m x^2 $$

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 $\left(\frac{x}{10}\right)^{n} \mathrm{~J}$. Therefore, we have: $$ -m x^2=\left(\frac{10}{x}\right)^{-n} $$

Because this is a loss of kinetic energy, we should consider the absolute value. Hence, $$ m x^2=\left(\frac{10}{x}\right)^{-n} $$

Substituting the given mass $m=10 \mathrm{~g}=0.01 \mathrm{~kg}$, we get: $$ 0.01 x^2=\left(\frac{10}{x}\right)^{-n} $$

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

Comparing the two sides, we can see that $n=1$.