11SCI - Mechanics
Finn Le Sueur
2024
The amount of energy transferred/transformed (gained/lost).
E.g. When you lift your backpack off the ground you are transferring some chemical potential energy in your muscles into gravitational potential energy in the backpack. You are doing work on your backpack. If the overall energy does not change, no work is done.
Work has the formula:
\[ \begin{aligned} work &= force \times distance \newline W &= F \times d \end{aligned} \]
and is measured in Joules (J) because it is defined as the amount of energy transferred/transformed.
Rearrange the formula \(W=F \times d\) to:
\[ \begin{aligned} W &= F \times d \newline F &= \frac{W}{d} \newline d &= \frac{W}{F} \end{aligned} \]
\[ \begin{aligned} K:& \cr U:& \cr F:& \cr S+S:& \end{aligned} \]
\[ \begin{aligned} W &= F \times d \newline W &= (m \times g) \times d \newline W &= (20 \times 10) \times 1 = 200J \end{aligned} \]
\[ \begin{aligned} W &= F \times d \newline W &= (m \times g) \times d \newline 605,000 &= (55 \times 10) \times d \newline \frac{605000}{55 \times 10} &= d = 1100m \end{aligned} \]
Complete and mark!
Compare and contrast these formula with the person sitting next to you. Write your thoughts in your book.
\[ \begin{aligned} W &= F \times d \newline E_{p} &= m \times g \times h \end{aligned} \]
\[ \begin{aligned} K:& \cr U:& \cr F:& \cr S+S:& \end{aligned} \]
\[ \begin{aligned} W &= F \times d \newline W &= (m \times g) \times d \newline W &= (62 \times 10) \times 46.2 = 28,644J \end{aligned} \]
We use the vertical distance, not the horizontal distance. Energy is not transferred/transformed when moving horizontally!
Get ready to Kahoot!
A rocket is launched with an acceleration of \(90ms^{-2}\). It has a mass of \(5kg\) and it reaches a height of 2000m. How much work did the rocket do to get the rocket to this height?
\[ \begin{aligned} K:& \cr U:& \cr F:& \cr S+S:& \end{aligned} \]
All energy is transformed from chemical potential to gravitational potential.
\[ \begin{aligned} W = E_{p} &= m \times g \times h \newline &= 5 \times 10 \times 2000 = 100,000J \end{aligned} \]
Complete Q33 then Q32 from your homework booklet.
\[ \begin{aligned} K:& \cr U:& \cr F:& \cr S+S:& \end{aligned} \]
Extra: Q28
\[ \begin{aligned} W = E_{p} &= m \times g \times h \newline &= 48 \times 10 \times 2 = 960J \end{aligned} \]
Because \(W=mgh\), if \(m\) increases, the work increases. Also if \(h\) increases, so does the work. Ian:
\[ \begin{aligned} W = E_{p} &= m \times g \times h \newline &= 52 \times 10 \times 5 = 2600J \end{aligned} \]
The rate at which energy is transferred/transformed.
The rate at which work is done.
\[ \begin{aligned} power &= \frac{work}{time} \newline P &= \frac{W}{t} \newline \end{aligned} \]
Write this formula so that \(W\) and \(t\) are the subject.
\[ \begin{aligned} K:& \cr U:& \cr F:& \cr S+S:& \end{aligned} \]
\[ \begin{aligned} P &= \frac{W}{t} \newline &= \frac{1,134,000}{25,200} \newline &= 45Js^{-1} = 45\frac{J}{s} = 45W \end{aligned} \]
\[ \begin{aligned} W &= F \times d \newline W &= m \times g \times h \end{aligned} \]
\[ \begin{aligned} P &= \frac{W}{t} \newline P &= \frac{m \times g \times h}{t} \newline 192 &= \frac{48 \times 10 \times 2}{t} \newline t &= \frac{48 \times 10 \times 2}{192} = 5s \end{aligned} \]
Complete and mark Q36 from your Mahi Kāinga booklet. Hint: find work first, and then power.
\[ \begin{aligned} K:& \cr U:& \cr F:& \cr S+S:& \end{aligned} \]
\[ \begin{aligned} W &= F \times d \newline &= (62 \times 10) \times 46.2 = 28,644J \end{aligned} \]
\[ \begin{aligned} P & = \frac{W}{t} \newline &= \frac{28644}{525} = 54.56Js^{-1} = 54.56\frac{J}{s} = 54.56W \end{aligned} \]