Work, Energy and Power

$?\frac{P_i=P_f}{m*40}=\frac{m}{2}v\text{'}+\frac{m}{2}*60$

v’ = 20 m/s

$K{E}_f=\frac{1}{2}\frac{m}{2}\left[v^2+60^2\right]$

$K{E}_f=\frac{1}{2}\frac{m}{2}\left[v^2+60^2\right]$

$\frac{\Delta KE}{KE}=\frac{\frac{1}{2}m\left(2000\right)-\frac{1}{2}m\left(1600\right)}{\frac{1}{2}m\left(1600\right)}$

$\therefore x=1$

Velocity of bob after collision,

v_{1 $=\sqrt[]{5gR}=\sqrt[]{5*10*0.5}=5m/s$}

From Conservation of Momentum,

$\frac{10}{1000}*v=\frac{-10}{1000}*100+1*5$  

->v = 400 m/s  

W_A = W_B

$\Rightarrow F_A*d*cos45=F_B*d*cos60$     

$\Rightarrow \frac{F_A}{F_B}=\frac{1}{\sqrt[]{2}}=\frac{1}{\sqrt[]{x}}$

=> x = 2

mgh = $ms\Delta T$

$\Rightarrow \Delta T=\frac{gh}{s}=\frac{10*63}{4200}=0.147°C$  

$\left[\lambda *3*g\right]*\frac{3}{2}-\left[\lambda *1*g\right]*\frac{1}{2}=K$         

=> $\lambda \Rightarrow \frac{mass}{length}$

=> $\lambda g\left(\frac{9}{2}-\frac{1}{2}\right)=K$

$\frac{3}{3}*10*4=K$

K = 40 J

$\frac{\Delta K}{K_i}*100=\frac{K_f-K_i}{K_i}*100$

$=\frac{\frac{P_f^2}{2m}-\frac{P_i^2}{2m}}{\frac{P_i^2}{2m}}*100=\frac{P_f^2-P_i^2}{P_i^2}*100$

$=\frac{{\left(1.2P_i\right)}^2-{\left(P_i\right)}^2}{P_i^2}*100=\frac{1.44P_i^2-P_i^2}{P_i^2}*100$

= 44%

The two main types of potential energy are:

1. Gravitational Potential Energy - Energy stored due to an object's position in a gravitational field (PE = mgh).
2. Elastic Potential Energy - Energy stored in deformed elastic objects like springs or stretched materials (PE = ½kx²).

The best way to find potential energy is to use the relationship 

${}_{}$ $U_f-U_i=-{\int }_i^f\overrightarrow{F}\cdot d\overrightarrow{r}$

and integrating the conservative force over the path.

For common cases, apply the standard potential energy formulas:

• PE = mgh for gravity
• PE = ½kx² for springs. 

Potential energy is best defined as a stored energy in a system. This energy is stored due to the configuration or position of the objects within a conservative force field. Potential energy in physics tells us that it's the capacity of the work done when the system is allowed to move from that position to the reference point.  

Com

m1v + m2 * 0 = m1v1 + m2v2

75v = $45\frac{v}{3}+50*10$ $\frac{}{}$

$\Rightarrow 75*\frac{2v}{3}=50*10$

$\Rightarrow v=\frac{3*50*10}{75*2}=10m/sec$