Question

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 \((10/x)^{-n}\)J. The value of n will be ____________

Answer 1

Correct Answer: 2

\(\begin{aligned} & \text { Given, } a=-2 x \\ & \Rightarrow \frac{v d v}{d x}=-2 x \\ & \Rightarrow v d v=-2 x d x \\ & \Rightarrow \int_{v_1}^{v_2} v d v=-2 \int_0^x x d x \\ & \Rightarrow \frac{v_2^2}{2}-\frac{v_1^2}{2}=-\frac{2 x^2}{2} \\ & \Rightarrow \frac{m v_1^2}{2}-\frac{m v_2^2}{2}=m x^2=\frac{10}{1000} x^2=10^{-2} x^2=\left(\frac{10}{x}\right)^{-2} \\ & \mathrm{n}=2 . \end{aligned}\)

Answer 2

Work-Energy Theorem and Retardation Problem 

Step 1: Work-Energy Theorem

The work-energy theorem states that the work done on a particle is equal to the change in its kinetic energy. In this case, the retardation force is doing negative work, leading to a loss of kinetic energy.

Step 2: Calculate Work Done

Given \( a = -2x \), where \( a \) is acceleration and \( x \) is displacement. We can write acceleration as \( a = v \frac{dv}{dx} \), so:

\( v \frac{dv}{dx} = -2x \)

\( v \ dv = -2x \ dx \)

Integrating both sides from initial to final states:

\( \int_{v_1}^{v_2} v \ dv = -2 \int_0^x x \ dx \)

\( \frac{v_2^2}{2} - \frac{v_1^2}{2} = -x^2 \)

Multiplying by mass \( m/2 \), gives

\(KE _2 - KE_1 = \frac{1}{2}m(v_2^2 - v_1^2) = -mx^2 \)

The change in kinetic energy is \( \frac{1}{2}mv_2^2 - \frac{1}{2}mv_1^2 = -mx^2 \). Given \( m = 10 \text{ g} = 0.01 \text{ kg} = 10 \times 10^{-3} \text{ kg} = 10^{-2} \text{ kg} \).

\( \Delta KE = -mx^2 = -(10 \times 10^{-3} \text{ kg})x^2 = -10^{-2}x^2 \text{ J} \)

\( \Delta KE = -x^2 \times 10^{-2} \text{ J} \)

Comparing this with the given expression for loss of kinetic energy, \( -x^{-n} \frac{x^2 \times 10}{\frac{2}{J}} \), we find \( n = 2 \).

Conclusion:

The value of \( n \) is \( \mathbf{2} \).