# Kinetic Energy

11SCI - Mechanics

Finn LeSueur

2019

# L20: Mahi Tuatahi (2016 Exam)

- Rhiana is riding a horse along New Brighton Beach.
**Each** of the horse’s hooves have a surface area of \(0.0044m^{2}\) which sink into the sand when the horse stops. The horse exerts a total pressure of \(200155Pa\).
**Calculate the weight of the horse.**
- Equation
- Substitute
- Solve

## Te Whakatika

A horse has four hooves, so the total surface area that the horse is exerting pressure through is \(A = 0.0044 \times 4 = 0.0176m^{2}\)

\[\begin{aligned}
P &= \frac{F}{A} \\
200155 &= \frac{F}{0.0176} \\
200155 \times 0.0176 &= F = 3522.728N
\end{aligned}\]
## Ngā Whāinga Ako

- Give the symbols and units for kinetic energy
- \(E_{k} = \frac{1}{2}mv^{2}\)

Write the date, the learning outcomes and title “Kinetic Energy” in your books.

## What is Kinetic Energy?

Kinetic energy is the energy that an object possesses due to its **velocity**!

## Calculating Kinetic Energy

Kinetic energy depends on the **mass** and **velocity** of an object.

\[\begin{aligned}
E_{k} &= \frac{1}{2} \times mass \times \text{velocity squared} \\
E_{k} &= \frac{1}{2} \times m \times v^{2}
\end{aligned}\]
### What does \(v^{2}\) mean?

- It means \(v \times v\)
- This means we can also write the equation like this, if you find it easier:

\[\begin{aligned}
& E_{k} = \frac{1}{2} \times m \times v \times v
\end{aligned}\]
### Pātai Tahi

Mr LeSueur rides his bike to work at \(32km/h\) (\(8.89ms^{-1}\)). Both he and his bike have a combined mass of \(80kg\). **Calculate his kinetic energy**.

#### Whakatika Tahi

We know \(m=80kg\) and \(v=8.89ms^{-1}\), and we are looking for \(E_{k}\).

\[\begin{aligned}
E_{k} = \frac{1}{2} \times 80 \times 8.89^{2} \\
E_{k} = 3161.284J
\end{aligned}\]
### Pātai Rua

Sophie is skiing down Upper Fascination at Mt Hutt, and is trying to go really fast. Her combined mass is \(60kg\) and she is moving at \(60km/h\) (\(16.67ms^{-1}\)). **Calculate her kinetic energy**.

#### Whakatika Tua

We know \(m=60kg\) and \(v=16.67ms^{-1}\), and we are looking for \(E_{k}\).

\[\begin{aligned}
E_{k} = \frac{1}{2} \times 60 \times 16.67^{2} \\
E_{k} = 8336.667J
\end{aligned}\]
## Whakamātau: Finding Your Kinetic Energy

Open the whakamātau document on Google Classroom!

**Homework:** Education Perfect due Monday 29th 11:25am
- sciPAD Page 46