Correct answer is (2)

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:

$\mathrm{\Delta }KE=W=\int F\cdot dx=\int (-2mx)dx=-m{x}^2$

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 ${\left(\frac{10}{x}\right)}^{-n}$ 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 $m=10\text{g}=0.01\text{kg}$, 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 $n=2$.

Correct answer is (67)

Spring energy at x = 10 cm,

${\mathrm{U}}_{\mathrm{i}}=\frac{1}{2}{\mathrm{kx}}_0^2$

Energy of the block at x = 10,

${\mathrm{K}}_{\mathrm{i}}=0$


Spring energy at x = 5 cm,

${\mathrm{U}}_{\mathrm{f}}=\frac{1}{2}\mathrm{k}{\left(\frac{{\mathrm{x}}_0}{2}\right)}^2$

Energy of the block at x = 5, (which is only kinetic energy, no potential energy of block presents as block is not moving in the vertical direction)

${\mathrm{K}}_{\mathrm{f}}=0.25\mathrm{J}$

Applying energy conservation law,

Initial energy of Spring + Initial energy of Block = Final energy of Spring + Final energy of Block

$\frac{1}{2}{\mathrm{kx}}_0^2+0=\frac{1}{2}\mathrm{k}\frac{{\mathrm{x}}_0^2}{4}+0.25$

$\frac{1}{2}{\mathrm{kx}}_0^2\frac{3}{4}=\frac{1}{4}$

$\frac{1}{2}\mathrm{k}\frac{3}{100}=1\Rightarrow \mathrm{k}=\frac{200}{3}\mathrm{N}/\mathrm{m}$

$=67\mathrm{N}/\mathrm{m}$

Correct answer is (7)

Given, $u=2\mathrm{m}/\mathrm{s}$

$\begin{array}{rl}& a=2\mathrm{m}/{\mathrm{s}}^2\\ \\ & s=6\mathrm{m}\\ \\ & v=?\\ \\ & v^2=u^2+2as\\ \\ & v^2=4+2\times 2\times 6\\ \\ & =28\end{array}$

So, $\mathrm{KE}=\frac{1}{2}m{v}^2=\frac{1}{2}\times 500\times 28\mathrm{J}$

$=7000\mathrm{J}$

$=7\mathrm{kJ}$

Correct answer is (300)

Displacement is $8^{\text{th }}$ sec.

${\mathrm{S}}_8=0+\frac{1}{2}\times 10\times (2\times 8-1)$

${\mathbf{S}}_8=5\times 15$

$\mathrm{\Delta }\mathrm{U}=0.4\times 10\times 5\times 15$

$\mathrm{\Delta }\mathrm{U}=20\times 15=300$

Correct answer is (2)

$\text{ Work done }=\mathrm{\Delta }\mathrm{K}.\mathrm{E}$

$\begin{array}{rl}& \therefore \int F\cdot dx=\frac{1}{2}m{v}^2=E\\ \\ & \therefore E={\int }_0^{x}2\cos (kx)dx\\ \\ & E=\frac{2}{k}[\sin kx{]}_0^x\\ \\ & =\frac{2}{k}\sin kx\\ \\ & =\frac{2\sin \theta }{k}\end{array}$