Ngā Whāinga Ako 🔗
- Be able to calculate gravitational potential energy ($E_{p} = mgh$)
- Be able to calculate kinetic energy ($E_{k} = \frac{1}{2}mv^{2}$)
- Be able to use the law of conservation of energy
Write the date and ngā whāinga ako in your book
What is Energy? 🔗
Energy is a quantity that must be transferred/transformed to do work.
- Pātai: What is energy measured in?
- Whakatika: Joules (J)
Work is the amount of energy trasferred or transformed (J).
The Different Forms of Energy 🔗
- Light
- Heat
- Sound
- Electrical
- Radiation
- Kinetic
- Nuclear potential
- Chemical potential
- Gravitational potential
- Elastic potential
Law of Conservation of Energy 🔗
Energy can neither be created nor destroyed, it can only be transformed or transferred.
This tells us that: the total energy in a system is always conserved. The system might be a collision/explosion, a beaker, Earth or the whole Universe!
Seeing it in Action 🔗
- For example, in this simulation, the skater will never go higher than they started.
- This is because they cannot get extra energy from the surroundings.
- In a frictionless world, they will also reach the same height because no energy is lost to the surroundings!
Kinetic Energy 🔗
The energy that a moving object has. Related to its velocity and mass.
$$ \begin{aligned} & E_{k} = \frac{1}{2}mv^{2} \cr & \text{m = mass of the moving object} \cr & \text{v = speed of the moving object} \end{aligned} $$
Gravitational Potential Energy 🔗
Energy an object has by being displaced from ground in a gravitational field. Related to its mass and height.
$$ \begin{aligned} & E_{p} = mg \Delta h \cr & \text{m = mass of the object} \cr & \text{g = acceleration due to gravity } 9.8ms^{-2} \downarrow \cr & \text{h = height of the object} \end{aligned} $$
Combining Gravitational and Kinetic Energy 🔗
When an object falls from a height, its gravitational potential energy is transformed into kinetic energy, but, the total energy in the system is constant.
In the real world some energy is lost due to friction as heat, light or sound. In the ideal world 100% of the energy is transformed.
In an ideal world we say:
$$ \begin{aligned} E_{total} &= E_{k} + E_{p} \cr E_{k} &= E_{p} && \text{they are equal} \cr \frac{1}{2}mv^{2} &= m g \Delta h && \text{substitute in the equations} \cr \end{aligned} $$
Pātai: The Sandbag 🔗
A bullet of mass $30g$ is fired with a speed of $400ms^{-1}$ into a sandbag. The sandbag has a mass of $10kg$ and is suspended by a rope so that it can swing.
- Find the kinetic energy of the bullet
- Find the height that the sandbag + bullet could rise
Te Whakatika 🔗
Step 1. Find kinetic energy of the bullet
$$ \begin{aligned} & E_{k} = \frac{1}{2}mv^{2} \cr & E_{k} = \frac{1}{2} \times 0.03 \times 400^{2} && \text{substitute values} \cr & E_{k} = 2,400J \end{aligned} $$
Step 2. Equate this with potential energy of sandbag & bullet
$$ \begin{aligned} E_{total} &= E_{k} + E_{p} \cr E_{k} &= E_{p} && \text{they are equal} \cr E_{k} &= m g \Delta h \cr 2400 &= 10.03 \times 9.8 \times \Delta h && \text{substitute values} \cr 2400 &= 98.294h \cr \Delta h &= \frac{2400}{98.294} \cr \Delta h &= 24.41m \end{aligned} $$
Whakawai/Practise 🔗
- Homework Booklet: Q67a, Q70
Ngā Whāinga Ako 🔗
- Be able to calculate the energy stored in a spring ($E_{p} = \frac{1}{2}kx^{2}$)
- Be able to use Hooke’s Law ($F=-kx$)
Write the date and ngā whāinga ako in your book
Elastic Potential Energy 🔗
A spring displaced from equilibrium will store some potential energy (to return to equilibrium).
$$ \begin{aligned} & E_{p} = \frac{1}{2}kx^{2} \cr & \text{k = spring constant} \cr & \text{x = spring compression/stretch (displacement)} \end{aligned} $$
The spring constant (k) is a measure of the stiffness of the spring.
Hooke’s Law 🔗
We can relate the displacement of a spring to its spring constant and the force required to create the displacement using Hooke’s Law.
$$ \begin{aligned} F &= -kx && \text{Force exerted by spring} \cr \end{aligned} $$
- The $-$ indicates that the restoring force and the displacement are in opposite directions.
- $F$: the force displacing the spring (Newtons)
- $x$: the displacement of the spring (meters)
- $k$: the spring constant ($Nm^{-1}$)
Pātai Rua (Q2) 🔗
Paris has a mass of $55kg$ and she is a spectator at a sports game. She steps onto a bench to get a good view. The bench is $4m$ long and it is displaced by $3mm$ in the middle when she stands on it.
- Calculate the spring constant of the bench. (M)
- Give correct SI units for the spring constant. (A)
- Calculate the elastic potential energy stored in the bench. (A)
Whakatika 🔗
- Calculate the spring constant of the bench + unit
$$ \begin{aligned} & k = \frac{F}{x} \cr & k = \frac{55 \times 9.8}{0.003} \cr & k = 179667Nm^{-1} \cr & k = 1.8\times10^{5}Nm^{-1} \end{aligned} $$
- Calculate the elastic potential energy stored in the bench. (A)
$$ \begin{aligned} & E_{p} = \frac{1}{2} k x^{2} \cr & E_{p} = \frac{1}{2} \times 1.8\times10^{5} \times 0.003^{2} \cr & E_{p} = 0.81J \end{aligned} $$
Pātai Toru (Q3): Aeroplane 🔗
A toy aeroplane ($500g$) is hanging at the end of a spring. The spring is $48.0cm$ long when hanging vertically. When the aeroplane is hung from the end of the spring, the length of spring becomes $80.0cm$.
- Calculate the spring constant. (M)
- Write a unit with your answer. (A)
- Calculate the energy stored in the spring when a second toy of mass $400g$ is also hung along with the aeroplane. (M)
- The $500g$ aeroplane is now hung on a stiffer spring, which has double the spring constant. Discuss how this affects the extension and the elastic potential energy in the spring. (E)
Whakatika 🔗
- Calculate the spring constant. (M)
$$ \begin{aligned} & k = \frac{F}{x} \cr & k = \frac{0.5 \times 9.8}{0.32} \cr & k = 15.31Nm^{-1} \end{aligned} $$
- Calculate the energy stored in the spring when a second toy of mass $400g$ is also hung along with the aeroplane. (M)
$$ \begin{aligned} & x = \frac{F}{k} \cr & x = \frac{0.9 \times 9.8}{15.31} \cr & x = 0.576m \end{aligned} $$
$$ \begin{aligned} & E_{p} = \frac{1}{2}kx^{2} \cr & E_{p} = \frac{1}{2} \times 15.31 \times 0.576^{2} \cr & E_{p} = 2.54J \end{aligned} $$
- The $500g$ aeroplane is now hung on a stiffer spring, which has double the spring constant. Discuss how this affects the extension and the elastic potential energy in the spring. (E)
$$ \begin{aligned} & x = \frac{F}{k} \cr & x = \frac{0.5 \times 9.8}{30.62} \cr & x = 0.16m \end{aligned} $$
It halves the amount that the spring extends, and reduces the amount of energy stored by a lot.
$$ \begin{aligned} & E_{p} = \frac{1}{2}kx^{2} \cr & E_{p} = \frac{1}{2} \times 30.62 \times 0.16^{2} \cr & E_{p} = 0.39J \end{aligned} $$
Whakawai/Practise 🔗
- Homework Booklet: 47-48
Mahi Tuatahi 🔗
- Give the equations for kinetic, gravitational potential and elastic potential energy.
- Give the name and formula for the law that you can use to relate force, spring constant and displacement.
- Lachie goes to a football game in a van with some of his teammates. It has suspension on each wheel. Lachie and his teammates weight $357kg$ in total and their weight is spread evenly across all four springs. The springs have a spring constant of $2.26 \times 10^{4}Nm^{-1}$. Calculate how much the car sinks down when they get into the car. (E)
- How much energy is stored in each spring if car sinks $0.12m$? (A)
Whakatika 🔗
- Give the equations for kinetic, gravitational potential and elastic potential energy.
$$ \begin{aligned} & E_{k} = \frac{1}{2}mv^{2} && \text{kinetic energy} \cr & E_{p} = m g \Delta h && \text{gravitational potential energy} \cr & E_{p} = \frac{1}{2} k x^{2} && \text{elastic potential energy} \end{aligned} $$
- Give the name and formula for the law that you can use to relate force, spring constant and displacement.
$$ \begin{aligned} & F = kx && \text{Hooke’s Law} \end{aligned} $$
- Calculate how much the car sinks down when they get into the car. (E)
Step 1: Weight per Spring
$$ \begin{aligned} & F = \frac{357 \times 9.8}{4} \cr & F = 874.65N \end{aligned} $$
Step 2: Displacement
$$ \begin{aligned} & F = kx && \text{Hooke’s Law} \cr & x = \frac{F}{k} \cr & x = \frac{874.65}{2.26 \times 10^{4}} \cr & x = 0.0387m \cr \end{aligned} $$
- How much energy is stored in each spring if they are compressed by $0.12m$? (A)
$$ \begin{aligned} & E_{p} = \frac{1}{2}kx^{2} \cr & E_{p} = \frac{1}{2} 2.26 \times 10^{4} \times 0.12^{2} \cr & E_{p} = 160J \end{aligned} $$
Ngā Whāinga Ako 🔗
- Be able to define & calculate work done ($W=Fd=mgh$)
- Be able to define & calculate power ($P=\frac{W}{t}$)
Write the date and ngā whāinga ako in your book
Work 🔗
The amount of energy transferred/transformed (Joules, J).
$$ \begin{aligned} & W = Fd \cr & work = force \times distance \end{aligned} $$
- Consider moving an object through a gravitational field
- You have to exert a force against a weight force, to lift it some distance
$$ \begin{aligned} W &= F_{w}d \cr W &= (m \times g) d \cr W &= mgh \end{aligned} $$
- Work is done only when energy is transferred or transformed.
- Work Done: Lifting an object and placing it on a shelf transfers energy to that object into the form of gravitational potential energy.
- No Work Done: Moving an object horizontally where it starts and finishes with $v=0ms^{-1}$.
- Work only depends on the start and finish position, the path does not matter (path independence).
Pātai Whā (Q4): Eddie Hall 🔗
In 2016 weightlifter Eddie Hall set a new (at the time) world record for heaviest deadlift of $500kg$. If he lifted the weights to a height of $1.25m$, how much work did Eddie do?
Whakatika 🔗
$$ \begin{aligned} & m=500kg, d=h=1.25m && \text{(K)} \cr & W=? && \text{(U)} \cr & W=Fd=mgh && \text{(F)} \cr & W=500\times9.81\times1.25= 6131.25J && \text{(S+S)} \end{aligned} $$
Power 🔗
The rate at which energy is transferred/transformed (the rate at which work is done).
$$ \begin{aligned} & P = \frac{W}{t} \cr & power = \frac{work}{time} \cr & power = \frac{Joules}{seconds} \cr & power = Js^{-1} && \text{also known as a Watt (W)} \end{aligned} $$
Pātai Rimu: Eddie Hall 🔗
If it took Eddie $7s$ to do $6125J$ of work on the weights, what power was he exerting?
Whakatika 🔗
$$ \begin{aligned} & W = 6125J, t=7s && \text{(K)} \cr & P = ? && \text{(U)} \cr & P = \frac{W}{t} && \text{(F)} \cr & P = \frac{6125}{7} = 875Js^{-1} && \text{(S+S)} \end{aligned} $$
Whakawai/Practise 🔗
- Homework Booklet: 65, 66b, 69, 68