Work, Energy and Power

$F=-\alpha x^2\Rightarrow \frac{mvdv}{dx}=-\alpha x^2$

$\Rightarrow {\displaystyle \underset{v_0}{\overset{0}{\int }}vdv=-\frac{\alpha }{m}{\displaystyle \underset{0}{\overset{x}{\int }}{x}^{2}dx\Rightarrow \frac{0^2}{2}-\frac{v_0^2}{2}=\frac{-\alpha }{3m}{x}^3}}$

$\Rightarrow \frac{-v_0^2}{2}=-\frac{\alpha }{3m}{x}^3\Rightarrow x={\left(\frac{3v_0^{2}m}{2\alpha }\right)}^{1/3}$

Using conservation of energy between initial and final position shown :

$\left[\frac{1}{2}\left(\frac{2}{5}m{a}^2\right){\left(\frac{V_0}{a}\right)}^2+\frac{1}{2}m{V}_0^2\right]+0=mg\mathcal{l}sin\theta$

$\Rightarrow \frac{7}{10}m{V}_0^2=mg\mathcal{l}sin\theta$

$\Rightarrow \mathcal{l}=\frac{7v_0^2}{10gsin\theta }$

At central point on screen, path difference is zero for all wavelength. So, central bright fringe is white and other fringes depend on wavelength as $\beta =\frac{\lambda D}{d}$ .

Therefore, other fringes will be coloured.

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

Formula for Power = F·v cos $$$\theta$

According to the question, force is perpendicular.

So $\theta$ = 90

and cos 90 = 0

Hence, the result turns out to be 0 i.e. no power generated inspite of force being applied.

In simple words,

Energy = total work done on an object.

Power = how fast the work is done on an object.

For example: Writing an article vs how fast you write the article. This sums up the exact meaning of both these terms.

No. Even if the power is higher, the energy can be inefficient. This happens when much of the energy generated gets wasted or isn't utilized as effectively as it should have been. In such cases, high power can still lead to inefficiency of energy.

Let m = 250 g = 0.25 kg

By Reduced mass method

$m_r=\frac{m_1{m}_2}{m_1+m_2}=\frac{mm}{m+m}=\frac{m}{2}$

By wet

$w_{SP}=\Delta K.E.$

$-\frac{1}{2}k{x}^2=0-\frac{1}{2}\left(\frac{m}{2}\right){\left(2v\right)}^2$

$\frac{2}{2}{x}^2=0.25v^2$

$x^2=0.25v^2$

$x=\frac{v}{2}$

$x=2t-1$

$v=\frac{dx}{dt}=2 m s^{-1}$

$P=F·v$

$=2*5=10 W$

No. Since kinetic energy is a scalar quantity, it only depends on speed of the body and not the direction. So if the direction of the body is changed but the speed remains unchanged, there won't be any effect on the kinetic energy. However, if changing the direction also changes the speed of the body, then kinetic energy of the body will also change.