Worked Examples for Energy


These worked examples will help to solidify your understanding towards the concept of conservation of energy and energy conversion.


Worked Example 1: Speeding Projectile

A projectile of mass 0.02 kg travels at a speed of $1200 \text{ m s}^{-1}$. Calculate its kinetic energy.

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$$\begin{aligned} E_{k} &= \frac{1}{2} mv^{2} \\ &= \frac{1}{2} (0.02)(1200)^{2} \\ &= 14400 \, \text{J} \end{aligned}$$


Worked Example 2: Lifting Object

An object with a mass of 5 kg is lifted vertically through a distance of 10 m at a constant speed. What is the gravitational potential energy gained by the object?

(Take the acceleration due to gravity to be $10 \text{ m s}^{-2}$)

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$$\begin{aligned} E_{gpe} &= mgh \\ &= (5)(10)(10) \\ &= 500 \, \text{J} \end{aligned}$$


Worked Example 3: Nail In A Plank

A nail is being hammered into a plank.

  1. What energy does a raised hammer possess?
  2. When it falls, what energy will the energy in part a. convert into?
  3. What is the subsequent energy used for?
  4. Are there any other forms of energy produced? If so, name them.
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  1. Gravitational potential energy
  2. Kinetic energy
  3. It is used to drive the nail into the plank
  4. Some of the gravitational potential energy is converted into sound and thermal energy


Worked Example 4: Energy Conversion of A Stone

A boy throws a stone into the air and catches it on the way down. State the energy conversions that take place.

Note: This will be a slight variation of the Case Study 2: Bouncing Ball.

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  1. Just after the stone leaves the boy’s hand, the stone has maximum kinetic energy and minimum gravitational potential energy.
  2. As it rises, it’s kinetic energy is gradually converted into gravitational potential energy.
  3. At the point of maximum height, the stone has only gravitational potential energy. At this point, the stone has stopped momentarily so the kinetic energy of the stone is zero.
  4. As it is falling down, the stone’s gravitational potential energy is gradually converted into kinetic energy.
  5. Just before the stone reaches the boy’s hand, it will have maximum kinetic energy and minimum gravitational potential energy. However, due to energy losses from friction/air resistance, the new kinetic energy of the stone will still be smaller than it’s initial kinetic energy.


Worked Example 5: Block on Frictionless Slope

block on slope

A block of mass 5 kg slides from rest (at the top of the slope) through a distance of 30 m down a frictionless slope. What is the kinetic energy of the block at the bottom of the slope?

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Note that the height of the slope is 5 m. At the top of the slope, the block will possess some gravitational potential energy.

The gravitational potential energy of the block at the top of the slope is given by:

$$\begin{aligned} E_{GPE} &= mgh \\ &= 5 \times 10 \times 5 \\ &= 250 \text{ J} \end{aligned}$$

As the block slides down the slope, the gravitational potential energy will be converted into kinetic energy. Hence, when the block reaches the bottom of the slope, ALL the gravitational potential energy will be converted into kinetic energy.

The kinetic energy of the block at the bottom of the slope will be 250 J.


Worked Example 6: Block on a rough Slope

block on slope

A block of mass 5 kg slides from rest (at the top of the slope) through a distance of 30 m down a rough slope. Considering that 30 J is dissipated as heat due to friction, what is the kinetic energy of the block at the bottom of the slope?

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Notice that the question is the same as the previous worked example except for the presence of friction.

Similarly, the gravitational potential energy of the block at the top of the slope will be 250 J.

As the block slides down the slope, the gravitational potential energy will be converted into kinetic energy AND heat (friction). Hence, when the block reaches the bottom of the slope, ALL the gravitational potential energy will be converted into kinetic energy AND heat (friction).

This means that:

$$\text{Gravitational Potential Energy} = \text{Kinetic Energy} + \text{Heat}$$

Given that 30 J is dissipated as heat, we have:

$$\begin{aligned} E_{kin} &= 250-30 \\ &= 220 \text{ J} \end{aligned}$$


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