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 :