Work and Energy • Topic 3 of 3

Conservation of Energy & Power

One of the most important ideas in all of science is the law of conservation of energy: energy can neither be created nor destroyed; it can only be transformed from one form to another. The total energy of a system (and of the whole universe) stays constant. When energy seems to disappear, it has actually changed into another form — often heat or sound that is hard to notice.

The clearest school example is a freely falling body. Imagine a ball dropped from a height $h$. At the top it is at rest, so its kinetic energy is zero and all its energy is potential, $E_p=mgh$. As it falls, its height decreases (so $E_p$ falls) while its speed increases (so $E_k$ rises). At every point the sum stays the same:

$E_p+E_k=\text{constant}$, i.e. $mgh+\frac{1}{2}mv^2=mgh_{\text{top}}$

Just before hitting the ground all the potential energy has become kinetic energy. A simple pendulum shows the same idea repeating: at the highest point of its swing it has maximum potential energy and zero kinetic energy, while at the lowest point it has maximum kinetic energy and minimum potential energy. Energy keeps shuttling between $E_p$ and $E_k$ (in real life friction and air resistance slowly convert some into heat, so the swing dies down).

Power tells us how fast work is done or how fast energy is transferred. Two cranes may lift the same load to the same height — doing the same work — but the one that does it in less time is more powerful. Power is defined as the rate of doing work:

$P=\frac{W}{t}$

where $W$ is the work done (or energy transferred) and $t$ is the time taken. The SI unit of power is the watt (W), named after James Watt. One watt is one joule of work done in one second, so $1\,\text{W}=1\,\text{J/s}$. Larger units are the kilowatt ($1\,\text{kW}=1000\,\text{W}$) and the megawatt ($1\,\text{MW}=10^6\,\text{W}$). The older unit horsepower is roughly $746\,\text{W}$.

For everyday electricity bills, the joule is too small, so we use a much larger commercial unit of energy: the kilowatt-hour, written $\text{kWh}$ and often called one ‘unit’ of electricity. One kilowatt-hour is the energy used by a 1 kW appliance running for 1 hour. Converting to joules: $1\,\text{kWh}=1000\,\text{W}\times3600\,\text{s}=3.6\times10^6\,\text{J}$. So a 1000 W heater used for one hour consumes exactly 1 kWh of energy.

Energy bar chart showing conservation of energy during a fallConservation of Energy: PE + KE = constantEnergy (J)TopPE maxMiddleBottomKE maxPEKETotal height of each bar (PE + KE) stays the same throughout the fall
1
Worked Example
A machine does 600 J of work in 5 s. Calculate its power.
Solution
  1. Given: $W=600\,\text{J}$, $t=5\,\text{s}$.
  2. $P=\frac{W}{t}$.
  3. $P=\frac{600}{5}=120\,\text{W}$.

Answer: Power $=120\,\text{W}$.

2
Worked Example
A ball of mass 0.5 kg is dropped from a height of 8 m. Find its speed just before it hits the ground using energy conservation. Take $g=10\,\text{m/s}^2$.
Solution
  1. At the top, $E_p=mgh=0.5\times10\times8=40\,\text{J}$, $E_k=0$.
  2. By conservation, all PE becomes KE at the bottom: $\frac{1}{2}mv^2=40$.
  3. $\frac{1}{2}\times0.5\times v^2=40\Rightarrow0.25v^2=40\Rightarrow v^2=160\Rightarrow v\approx12.6\,\text{m/s}$.

Answer: Speed $\approx12.6\,\text{m/s}$.

3
Worked Example
An electric bulb is rated 60 W. How much energy does it use in 2 hours? Express the answer in kWh and in joules.
Solution
  1. Power $=60\,\text{W}=0.06\,\text{kW}$; time $=2\,\text{h}$.
  2. Energy $=P\times t=0.06\times2=0.12\,\text{kWh}$.
  3. In joules: $0.12\times3.6\times10^6=4.32\times10^5\,\text{J}$.

Answer: $0.12\,\text{kWh}=4.32\times10^5\,\text{J}$.

4
Worked Example
A pump raises 200 kg of water to a height of 10 m in 40 s. Find the power of the pump. Take $g=10\,\text{m/s}^2$.
Solution
  1. Work done $=mgh=200\times10\times10=20000\,\text{J}$.
  2. $P=\frac{W}{t}=\frac{20000}{40}$.
  3. $P=500\,\text{W}$.

Answer: Power of the pump $=500\,\text{W}$.

5
Worked Example
At the highest point of a pendulum swing, the bob has 12 J of potential energy and is momentarily at rest. Assuming no friction, what is its kinetic energy at the lowest point?
Solution
  1. At the highest point: $E_p=12\,\text{J}$, $E_k=0$.
  2. By conservation of energy, total mechanical energy stays at 12 J.
  3. At the lowest point $E_p=0$, so $E_k=12\,\text{J}$.

Answer: Kinetic energy at the lowest point $=12\,\text{J}$.

6
Worked Example
A household uses a 2 kW geyser for 1.5 hours daily. If electricity costs Rs 8 per unit (kWh), find the daily cost.
Solution
  1. Energy per day $=P\times t=2\,\text{kW}\times1.5\,\text{h}=3\,\text{kWh}$.
  2. Cost $=$ units $\times$ rate $=3\times8$.
  3. Cost $=\text{Rs }24$.

Answer: Daily cost $=\text{Rs }24$.

Key Points

  • Law of conservation of energy: energy is neither created nor destroyed, only transformed; total energy stays constant.
  • In free fall and in a pendulum, energy converts between potential and kinetic so that $E_p+E_k=$ constant.
  • Power is the rate of doing work: $P=\frac{W}{t}$.
  • The SI unit of power is the watt (W), where $1\,\text{W}=1\,\text{J/s}$; $1\,\text{kW}=1000\,\text{W}$.
  • The commercial unit of energy is the kilowatt-hour: $1\,\text{kWh}=3.6\times10^6\,\text{J}$.
Tap an option to check your answer0 / 4
Q1.According to the law of conservation of energy, energy can be
Explanation: Energy is neither created nor destroyed; it only changes from one form to another.
Q2.The SI unit of power is the
Explanation: Power is measured in watts: $1\,\text{W}=1\,\text{J/s}$.
Q3.At the lowest point of a frictionless pendulum swing, the bob has maximum
Explanation: At the lowest point PE is minimum and KE is maximum.
Q4.One kilowatt-hour ($1\,\text{kWh}$) equals
Explanation: $1\,\text{kWh}=1000\,\text{W}\times3600\,\text{s}=3.6\times10^6\,\text{J}$.