Physics

Yes, all objects have inertia. It does not depend on whether they are moving or not. Inertia is an object's inherent resistance to any change in its state of motion.

For an object at rest, its inertia is its tendency to remain at rest. A force is required to overcome this inertia and set the object in motion. 

The amount of inertia an object has is determined by its mass. The more massive an object is, the more it resists a change in its state of motion. We can also say that this heavier object has greater inertia.

The law of inertia, also known as Newton's first law of motion, states that an object at rest will stay at rest. Likewise, an object in motion will remain in motion with the same speed and in the same direction unless there is a net external force. 

Now in both ideal states, the net external force on an object is zero.

From the perspective of classical mechanics, there isn't a significant distinction between rest and uniform motion. They can be seen as the same state of motion viewed from different reference frames. An object is considered to be in a state of equilibrium whether it is at rest or in uniform motion.

$A_{max}=A_c+A_m$

$A_{min}=A_c-A_m$

$\frac{A_{min}}{A_{max}}=\frac{A_c-A_m}{A_c+A_m}=\frac{250-150}{250+150}$

$\Rightarrow \frac{100}{400}=\frac{50}{200}$  

By work energy theorem

Work done = change in K.E.

Work done by friction work done by spring

$=0-\frac{1}{2}m{V}^2$              

As 90% of K.E. is losed by friction so that

$-\frac{90}{100}\left(\frac{1}{2}m{V}^2\right)-\frac{1}{2}K{x}^2=-\frac{1}{2}m{V}^2$                

-K -> -16 * 10⁵

K = 16 * 10⁵

$l=\frac{m{\mathcal{l}}^2}{3}$  

Energy conservation Low

$mg\mathcal{l}=\frac{1}{2}\frac{m{\mathcal{l}}^2}{3}{\omega }^2.....\left(i\right)$                

And speed V =    $\omega r=\omega \mathcal{l}$

then    $V=\sqrt[]{6g\mathcal{l}}=\sqrt[]{6*10*.6}$

->6 m/s

for smooth surface

$a=gsin30°=\frac{g}{2}$

$S_1=ut+\frac{1}{2}a{t}^2$         

$S_1=\frac{1}{2}\frac{g}{2}{t}^2=\frac{g}{4}{t}^2.......\left(i\right)$               

for rough Surface

$S=\frac{1}{2}\frac{g}{2}\left(1-\mu \sqrt[]{3}\right){\alpha }^2{t}^2........\left(ii\right)$               

By (i) and (ii)

$\mu =\frac{1}{\sqrt[]{3}}\left(\frac{{\alpha }^2-1}{{\alpha }^2}\right)x=\sqrt[]{3}$