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:

${\begin{matrix}U&=&E_{P}&+&E_{K}\\U&=&mgh&+&{\frac {1}{2}}mv^{2}\end{matrix}}$

(7.1)

Now,

IN THE ABSENCE OF FRICTION
Mechanical energy is conserved
${\begin{matrix}U_{initial}&=&U_{final}\\{\frac {1}{2}}mv_{i}^{2}+mgh_{i}&=&{\frac {1}{2}}mv_{f}^{2}+mgh_{f}\end{matrix}}$

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

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 (v_A = 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 (U_A) and at point B (U_B). 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 v_A, but what is h_A? Note that if we let h_B = 0 then h_A = 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 U_B = U_A. Therefore U_B = 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 h_B = 0. This means that

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