Worked Examples-Calculation Part 2 - Work, Energy and Power

Considered Formula;

$$T.E = K.E + P.E$$ $$\qquad \text {Where}~~T.E \to Total~energy,\quad K.E \to Kinetic~Energy,\\\qquad P.E \to Potential~Energy$$

Example 1

A stone of mass 2kg is released from a height of 2m above the ground. Find

a)    Total energy

b)    Potential energy at height of 0.5m

c)     Kinetic energy at height of 0.5m

d)     Velocity acquired at 0.5m

(Acceleration due to gravity (g) = 10 m/s²)

Solution

Data given

Mass (m) = 2 Kg

Maximum Height (h_max) = 2 m

Acceleration due to gravity (g) = 10 m/s²

a)        Total energy (T.E) = ?

$$~~~~~~~~~~~~~~~~~~~~~~~~~~~~T.E = P.E_{max} \quad \Rightarrow P.E_{max} \to \text {Maximum Potential energy}$$ $$~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~ = mgh_{max}$$ $$~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~ = 2 \times 10 \times 2$$ $$~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~ = 40~Joules$$

          The total energy = 40 Joules

b)        Potential energy at height of 0.5m (P.E) = ?

                 Height (h) = 0.5 m

                 Acceleration due to gravity (g) = 10 m/s²

$$~~~~~~~~~~~~~~~~~~~~~~~~~~~~P.E = mgh$$ $$~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~ = 2 \times 10 \times 0.5$$ $$~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~ = 10~Joules$$

           Potential energy at height of 0.5m = 10 Joules

c)        Kinetic energy at height of 0.5m (K.E) = ?

                 Potential energy at height of 0.5m (P.E) = 10 J

                 Total energy (T.E) = 40 J

$$ \qquad ~~~~~~~~~~~~~~~~~~~~~~~~~~~~T.E = K.E + P.E$$ $$ \qquad ~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~40 = K.E + 10$$ $$ \qquad ~~~~~~~~~~~~~~~~~~~~~~~~~~~~K.E = 40 - 10$$ $$ \qquad \qquad~~~~~~~~~~~~~~~~~~~~~~~~~~~~~= 30 J$$

           Kinetic energy at height of 0.5m = 30 Joules

d)        Velocity at height of 0.5m (v) = ?

                 Kinetic energy at height of 0.5m (K.E) = 30 J

$$~~~~~~~~~~~~~~~~~~~~~~~~~~~~K.E = \frac {1}{2}mv^2$$ $$~~~~~~~~~~~~~~~~~~~~~~~~~~~~30 = \frac {1}{2} \times 2 \times v^2$$ $$~~~~~~~~~~~~~~~~~~~~~~~~~~~~30 = v^2$$ $$~~~~~~~~~~~~~~~~~~~~~~~~~~~~v = \sqrt {30} m/s$$

           Velocity at height of 0.5m = √30 m/s

Example 2

A ball of mass 0.2 kg is dropped from a height of 20 m. On impact with the ground it loses 30 J of energy. Calculate the height which it reaches on the rebound. (Acceleration due to gravity (g) = 10 m/s²)

Solution

⇛Before rebound

Data given

Mass (m)= 0.2 Kg

Height (h) = 20 m

Accelaration due to gravity (g) =10 m/s²

Total energy (T.E) = ?

$$~~~~~~~~~~~~~~~~~~~~~~~~~~~~T.E = P.E_{max} \quad \Rightarrow P.E_{max} \to \text {Maximum Potential energy}$$ $$~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~ = mgh_{max}$$ $$~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~ = 0.2 \times 10 \times 20$$ $$~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~ = 40~Joules$$

The Total energy = 40 Joules

⇛After rebound

Potential energy at the second maximum height (P.E_h) = ?

Energy lost (E_lost) = 30 Joules

$$ \qquad ~~~~~~~~~~~~~~~~~~~~~~~~~~~~T.E = P.E_h + E_{lost}$$ $$ \qquad ~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~40 = P.E_h + 30$$ $$ \qquad ~~~~~~~~~~~~~~~~~~~~~~~~~~~~P.E_h = 40 - 30$$ $$ \qquad \qquad~~~~~~~~~~~~~~~~~~~~~~~~~~~~~= 10 J$$

The potential energy at the maximum height after rebound = 10 Joules

From

$$~~~~~~~~~~~~~~~~~~~~~~~~~~~~P.E_h = mgh$$ $$~~~~~~~~~~~~~~~~~~~~~~~~~~~~10 = 0.2 \times 10 \times h$$ $$~~~~~~~~~~~~~~~~~~~~~~~~~~~~h = \frac {10}{2}mgh$$ $$~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~ = 5~m$$ $$~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~ = 5~m$$

The height which it reaches on the rebound = 5 m