Conservation of Mechanical Energy

What is conservation of mechanical energy ?

• In an isolated system where only conservative forces cause changes, the kinetic energy and potential energy can change , but their sum , the mechanical energy E of the system , cannot change .

• Theory : If the kinetic energy is K and the potential energy is U of an object then,
  $K_{i}$ + $U_{i}$ = $K_{f}$ + $U_{f}$
  where ; the subscripts i and f refer to the initial and final states of the system.
  - Proof : When a conservative force does work W on an object within the system , that force transfers energy between kinetic energy K of the object and potential energy U of the system . From Work-Kinetic energy theorem , we can write ,
  $\Delta$ K = $K_{f}$ - $K_{i}$ = W
  And from Work-Potential energy theorem , we can write,
  $\Delta$ U = - W
  Now , combining this two equation , we find that ,
  $\Delta$ K = - $\Delta$ U
  We can rewrite this equation as ,
  $K_{f}$ - $K_{i}$ = - ( $U_{f}$ - $U_{i}$ )
  so, $K_{f}$ - $K_{i}$ = $U_{i}$ - $U_{f}$
  By rearranging this equation we find that ,
  $K_{i}$ + $U_{i}$ = $K_{f}$ + $U_{f}$
  

• Example : A boy of mass 50 kg is released from rest at the top of a water slide , at height h=10 m above the bottom of the slide . Assuming that the slide is frictionless because of the water on it . Find the boy's speed at the bottom of the slide.
• Answer : We have ,
  $K_{i}$ + $U_{i}$ = $K_{f}$ + $U_{f}$
  or, $\frac{1}{2}$m$v_{f}^{2}$ + mg$h_{f}$= $\frac{1}{2}$m$v_{i}^{2}$ + mg$h_{i}$
  Dividing by m and rearranging yield ,
  $v_{f}^{2}$ = $v_{i}^{2}$ + 2g( $h_{i}$ - $h_{f}$ )
  Putting $v_{i}$ = 0 and ( $h_{i}$ - $h_{f}$ ) = h leads to ,
  $v_{f}$ = $\sqrt{2gh}$
  = $\sqrt{2 \times 9.8 ms^{-2} \times 10 m}$
  = 14 m/s
• Try it yourself : A block of mass m = 2.0 kg is dropped from height h = 40 cm onto a spring of spring constant k= 1960 N/m . Find the maximum distance , the spring is compressed ? Note that , for a spring , U = $\frac{1}{2}$k$x^{2}$ .
• Experiments :