Work, Energy and Power

P = Fv = (ma)v = m (dv/dt)v
P dt = mv dv
∫P dt = ∫mv dv
Pt = ½mv²
v = √ (2Pt/m)
dx/dt = √ (2P/m) t¹/²
x = √ (2P/m) ∫t¹/² dt = √ (2P/m) * (2/3)t³/²
Squaring to match options:
x² = (8P/9m) t³
This does not match. Let's re-examine the options.
Position x is proportional to t³/².
x = (8P/9m)¹/² t³/²

W = ∫Fxdy
W = ∫ (5y + 20)dy from 0 to 10
W = [5y²/2 + 20y] from 0 to 10
W = (5 (10)²/2 + 20 (10) - 0 = 250 + 200 = 450 J

Using conversation of momentum in direction perpendicular to the original direction of motion,

mv₁ sin 30° = mv₂ sin 30°

Yes. Mechanical energy is the sum of both kinetic and potential energies. Kinetic energy cannot be a negative value since mass is always positive and velocity square leads to a positive value. However, potential energy can be negative depending on the reference point.

Not in all cases. This happens due to the interruption of non conservative forces like friction, air resistance, etc. which are responsible for decreasing the mechanical energy by converting it into other forms of energy such as heat, noise, etc.

Conservation of Mechanical Energy is linked to newton's second law of motion (F = ma) in combination with work energy theorem. The work energy theorem states that total work done on an object equals to change in kinetic energy. This theorem is used to prove that the sum of kinetic energy and potential energy remains constant.

A -> m_A = 1 kg, P_A = P

B ->  m_B = 2 kg,   P_B = P

$?kE=\frac{P^2}{2m}$

$\therefore \frac{k{E}_A}{k{E}_B}=\frac{\left(\frac{P^2}{2mA}\right)}{\left(\frac{P^2}{2m_B}\right)}=\frac{m_B}{m_A}=\frac{2}{1}$

Kindly consider the following figure

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

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

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

Using conservation of energy between initial and final position shown :

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

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

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