## Akoranga 37 Mahi Tuatahi

A ball is thrown upwards with an initial speed of $161.3km/hr$ ($44.8ms^{-1}$).

- How long does it take for the ball to reach its highest point?
- How high does the ball rise?

**Remember:** Knowns, Unknowns, Formula, Substitute, Solve

### Whakatika 1

__How long does it take for the ball to reach its highest point?__

Think in the $y-direction$!

\begin{aligned} & v_{i} = 44.8ms^{-1}, v_{f} = 0ms^{-1}, a = -9.8ms^{-2} && \text{(K)} \newline & t = ? && \text{(U)} \newline & v_{f} = v_{i} + at && \text{(F)} \newline & 0 = 44.8 + (-9.8)t && \text{(S+S)} \newline & \frac{-44.8}{-9.8} = t = 4.57s \end{aligned}

### Whakatika 2

__How high does the ball rise?__

Think in the $y-direction$!

\begin{aligned} & v_{i} = 44.8ms^{-1}, v_{f} = 0ms^{-1}, a = -9.8ms^{-2} && \text{(K)} \newline & d = ? && \text{(U)} \newline & v_{f}^{2} = v_{i}^{2} + 2ad && \text{(F)} \newline & 0^{2} = 44.8^{2} + 2(-9.8)d && \text{(S+S)} \newline & \frac{-2007.04}{2\times-9.8} = d = 102.4m \end{aligned}

## 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)

### 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

transformedortransferred.

This tells us that: **the total energy in the 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!

### What Energy Do We Care About?

- Kinetic
- Gravitational potential
- Elastic potential

## Kinetic Energy

The energy that a moving object has. Related to its velocity and mass.

\begin{aligned} & E_{k} = \frac{1}{2}mv^{2} \newline & \text{m = mass of the moving object} \newline & \text{v = speed of the moving object} \end{aligned}

## Gravitational Potential Energy

Energy an object has by being displaced from

groundin a gravitational field. Related to its mass and height.

\begin{aligned} & E_{p} = mg \Delta h \newline & \text{m = mass of the object} \newline & \text{g = acceleration due to gravity } 9.8ms^{-2} \downarrow \newline & \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.

Therefore when comparing an object at the top of its fall, to the bottom of its fall we can say:

\begin{aligned} & E_{total} = E_{k} + E_{p} \newline & E_{k} = E_{p} && \text{they are equal} \newline & \frac{1}{2}mv^{2} = m g \Delta h && \text{substitute in the equations} \newline \end{aligned}

### Pātai Tahi (Q1): 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.

Calculate the maximum height that the sandbag rises as it recoils with the bullet lodged inside.

#### Whakatika Tahi

**Step 1.** Find kinetic energy of the bullet

\begin{aligned} & E_{k} = \frac{1}{2}mv^{2} \newline & E_{k} = \frac{1}{2} \times 0.03 \times 400^{2} && \text{substitute values} \newline & E_{k} = 2,400J \end{aligned}

**Step 2.** Equate this with potential energy of sandbag & bullet

\begin{aligned} E_{total} &= E_{k} + E_{p} \newline E_{k} &= E_{p} && \text{they are equal} \newline E_{k} &= m g \Delta h \newline 2400 &= 10.03 \times 9.8 \times h && \text{substitute values} \newline 2400 &= 98.294h \newline \Delta h &= \frac{2400}{98.294} \newline \Delta h &= 24.41m \end{aligned}

### Whakawai/Practise

- Homework Booklet: Q70, Q66a, Q67a
- Textbook:

## Akoranga 38 Mahi Tuatahi

- What force do we know is
**not**acting due to the cars movement? - Therefore, what three forces
**are**acting? - Draw a force diagram illustrating these forces and their relative magnitude. Ensure you label them!

### Whakatika

- What force do we know is
**not**acting due to the cars movement?

_Thrust, because it is not accelerating._ - Therefore, what three forces
**are**acting?

_Weight, friction and support._ - Draw a force diagram illustrating these forces and their relative magnitude. Ensure you label them!

## 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} \newline & \text{k = spring constant} \newline & \text{x = spring compression/stretch (displacement)} \end{aligned}

- The spring constant is a measure of the stiffness of the spring.
- Low constant $\rightarrow$ easy to displace
- High constant $\rightarrow$ difficult to displace

## 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} \newline F &= kx && \text{Force creating the displacement} \newline \end{aligned}

- $F$: 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} \newline & k = \frac{55 \times 9.8}{0.003} \newline & k = 179667Nm^{-1} \newline & 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} \newline & E_{p} = \frac{1}{2} \times 1.8\times10^{5} \times 0.003^{2} \newline & 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} \newline & k = \frac{0.5 \times 9.8}{0.32} \newline & 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} \newline & x = \frac{0.9 \times 9.8}{15.31} \newline & x = 0.576m \end{aligned}

\begin{aligned} & E_{p} = \frac{1}{2}kx^{2} \newline & E_{p} = \frac{1}{2} \times 15.31 \times 0.576^{2} \newline & 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} \newline & x = \frac{0.5 \times 9.8}{30.62} \newline & 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} \newline & E_{p} = \frac{1}{2} \times 30.62 \times 0.16^{2} \newline & E_{p} = 0.39J \end{aligned}

### Whakawai/Practise

- Homework Booklet: 47-48
- Textbook:

## Akoranga 39 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 is going to football in the weekend. The van he rides in with some of his teammates 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} \newline & E_{p} = m g \Delta h && \text{gravitational potential energy} \newline & 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} \newline & F = 874.65N \end{aligned}

Step 2: Displacement

\begin{aligned} & F = kx && \text{Hooke’s Law} \newline & x = \frac{F}{k} \newline & x = \frac{874.65}{2.26 \times 10^{4}} \newline & x = 0.0387m \newline \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} \newline & E_{p} = \frac{1}{2} 2.26 \times 10^{4} \times 0.12^{2} \newline & 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).

One joule of work is done when a force of one newton moves an object one meter.

\begin{aligned} & W = Fd \newline & 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 \newline W &= (m \times g) d \newline 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)} \newline & W=? && \text{(U)} \newline & W=Fd=mgh && \text{(F)} \newline & 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} \newline & power = \frac{work}{time} \newline & power = \frac{Joules}{seconds} \newline & 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)} \newline & P = ? && \text{(U)} \newline & P = \frac{W}{t} && \text{(F)} \newline & P = \frac{6125}{7} = 875Js^{-1} && \text{(S+S)} \end{aligned}

### Whakawai/Practise

- Homework Booklet: 65, 66b, 69, 68
- Textbook: