The conservation of energy involving kinetic and potential energy
Individually, kinetic and potential energy can be calculated.
\({E_k} = \frac{1}{2}m{v^2}\)
and
\({E_p} = mgh\)
But when a mass is dropped from a height – where it has potential energy – what happens to that energy?
Or when a rocket shoots upwards with kinetic energy, what happens to the energy as the rocket gains altitude?
Due to the previously mentioned conservation of energy, one form of energy is transformed into another. Unless it is stated that there are any other energy losses or forces involved, the transform of potential energy to kinetic energy can be assumed as 100%. For example, for objects dropped through the air, or launched into the air, all of the \(E_{p}\) is transformed into \(E_{k}\), or all of the \(E_{k}\) is transformed into \(E_{p}\). This means that any air resistance is being ignored because its effect is minimal.
Example one
Question
A \(5kg\) gargoyle \(20 m\) high on a building becomes dislodged and falls to the ground.
At what speed does it reach the ground?
\(Starting\,potential\,energy = mgh\)
\(= 5 \times 9.8 \times 20\)
\(= 980J\)
\(Final\,kinetic\,energy = 980J\)
\((using\,the\,conservation\,of\,energy)\)
To find the final speed of the gargoyle use the kinetic energy formula.
\({E_k} = \frac{1}{2}m{v^2}\)
\(980 = \frac{1}{2} \times 5 \times {v^2}\)
\({v^2} = 980 \div (\frac{1}{2} \times 5)\)
\({v^2} = 980 \div 2.5\)
\({v^2} = 392\)
\(v = \sqrt {392}\)
\(v = 19.8\)
The gargoyle would reach the ground at 19.8 \(m\,s^{-1}\)
Example two
Question
A \(2.5kg\) rocket is launched vertically at \(30m\,s^{-1}\). Using the conservation of energy, what height will the rocket achieve?
\(Starting\,kinetic\,energy = \frac{1}{2}m{v^2}\)
\(= \frac{1}{2} \times 2.5 \times {30^2}\)
\(= 1125J\)
Because of the conservation of energy, all of this kinetic energy is transformed into potential energy as the rocket gains height.
\(Final\,potential\,energy = 1125J\)
\(Potential\,energy = mgh\)
\(1125 = 2.5 \times 9.8 \times h\)
\(1125 = 24.5 \times h\)
\(24.5 \times h = 1125\)
\(h=\frac{1125}{2\cdot 5\times9\cdot8}\)
\(h=45\cdot9\)
Height attained by rocket is \(45.9m\).
Example three
Question
A ball of mass \(0·3kg\) is thrown vertically upwards, reaches a maximum height of \(8 m\), then falls back towards the ground.
a) Calculate the potential energy of the ball when it reaches the maximum height.
b) As the ball returns to the ground, calculate the new potential energy of the ball when it falls to a height of \(5m\).
c) i) How much potential energy has the ball lost?
ii) Explain what has happened to this lost potential energy.
d) Calculate the speed of the ball at the height of \(5m\).
e) State any assumptions made in your answers.
a) \(E_{p}=mgh=0\cdot 3\,\times9\cdot 8\,\times8=23\cdot5\,J \)
b) \(E_{p}=mgh=0\cdot3\,\times9\cdot8\,\times5=14\cdot7\,J \)
c) i) \(lost\,E_{p}=23\cdot5\,-14\cdot7=8\cdot8\,J\)
ii) The lost potential energy has been transformed into the kinetic energy of the ball.
d) \(loss\,of\,E_{p}=E_{k}=\,_{1/2}mv^{2}=8\cdot8\,J \)
\(E_{k}=\,_{1/2}mv^{2} \)
\(8\cdot8=\,_{1/2}\times\,0\cdot3\,\times\,v^{2} \)
\(v=\sqrt{\frac{8\cdot8}{_{1/2}\,\times0\cdot3}}\)
\(=7\cdot7\,m\,s^{-1} \)
e) This assumes that air resistance is negligible.
