FHSST Physics/Work and Energy/Mechanical Energy and Energy Conservation

From Wikibooks, open books for an open world
< FHSST Physics‎ | Work and Energy
Jump to: navigation, search
The Free High School Science Texts: A Textbook for High School Students Studying Physics
Main Page - << Previous Chapter (Momentum) - Next Chapter (Collisions and Explosions) >>
Work and Energy
Definition - Work - Energy - Mechanical Energy and Energy Conservation - Important Quantities, Equations, and Concepts - Sources

Mechanical Energy and Energy Conservation[edit]

Kinetic energy and potential energy are together referred to as mechanical energy. The total mechanical energy (U) of an object is then the sum of its kinetic and potential energies:

\begin{matrix}U&=&E_P+E_K  \\U&=& mgh +\frac{1}{2}mv^2\end{matrix}

(7.1)

Now,

IN THE ABSENCE OF FRICTION
Mechanical energy is conserved
U_{before} = U_{after}
\frac{1}{2}mv_{0}^{2}+mgh_{0}=\frac{1}{2}mv_{f}^{2}+mgh_{f}

This principle of conservation of mechanical energy can be a very powerful tool for solving physics problems. However, in the presence of friction some of the mechanical energy is lost:

IN THE PRESENCE OF FRICTION
Mechanical energy is not conserved
(The mechanical energy lost is equal to the work done against friction)
\Delta U = U_{before} - U_{after} = \mathrm{Work\ Done\ Against\ Friction}

Worked Example 45 Using Mechanical Energy Conservation[edit]

Question: A 2kg metal ball is suspended from a rope. If it is released from point A and swings down to the point B (the bottom of its arc) what is its velocity at point B?

Fhsst wrkeng2.png

Answer:

Step 1 : Analyse the question to determine what information is provided

  • The mass of the metal ball is m = 2kg
  • The change in height going from point A to point B is h = 0.5m
  • The ball is released from point A so the velocity at point A is zero (vA = 0m/s).

These are in the correct units so we do not have to worry about unit conversions.


Step 2 : Analyse the question to determine what is being asked

  • Find the velocity of the metal ball at point B.

Step 3 : Determine the Mechanical Energy at A and B

To solve this problem we use conservation of mechanical energy as there is no friction. Since mechanical energy is conserved,

\begin{matrix}U_A=U_B\end{matrix}

Therefore we need to know the mechanical energy of the ball at point A (UA) and at point B (UB). The mechanical energy at point A is

\begin{matrix}U_A = mgh_A + \frac{1}{2}m(v_A)^2\end{matrix}

We already know m, g and vA, but what is hA? Note that if we let hB = 0 then hA = 0.5m as A is 0.5m above B. In problems you are always free to choose a line corresponding to h = 0. In this example the most obvious choice is to make point B correspond to h = 0.

Now we have,

\begin{matrix}U_A &=& (2kg)\left(10\frac{m}{s^2}\right)(0.5m)+\frac{1}{2}(2kg)(0)^2\\&=& 10\ J\end{matrix}

As already stated UB = UA. Therefore UB = 10J, but using the definition of mechanical energy

\begin{matrix}U_B &=& mgh_B + \frac{1}{2}m(v_B)^2\\&=& \frac{1}{2}m(v_B)^2\end{matrix}

because hB = 0. This means that

\begin{matrix}10J&=&\frac{1}{2}(2kg)(v_B)^2\\(v_B)^2 &=& 10\frac{J}{kg}\\v_B &=& \sqrt{10}\frac{m}{s}\end{matrix}