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

=Mechanical Energy and Energy Conservation=

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:

Now,



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:



Worked Example 45 Using Mechanical Energy Conservation
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?



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

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