12PHYS - Mechanics
Finn Le Sueur
2024
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Energy is a quantity that must be transferred/transformed to do work.
Work is the amount of energy trasferred or transformed (J).
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!
The energy that a moving object has. Related to its velocity and mass.
\[ \begin{aligned} & E_{k} = \frac{1}{2}mv^{2} \cr & \text{m = mass of the moving object} \cr & \text{v = speed of the moving object} \end{aligned} \]
Energy an object has by being displaced from ground in a gravitational field. Related to its mass and height.
\[ \begin{aligned} & E_{p} = mg \Delta h \cr & \text{m = mass of the object} \cr & \text{g = acceleration due to gravity } 9.8ms^{-2} \downarrow \cr & \text{h = height of the object} \end{aligned} \]
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:
\[ \begin{aligned} E_{total} &= E_{k} + E_{p} \cr E_{k} &= E_{p} && \text{they are equal} \cr \frac{1}{2}mv^{2} &= m g \Delta h && \text{substitute in the equations} \cr \end{aligned} \]
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.
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
\[ \begin{aligned} E_{total} &= E_{k} + E_{p} \cr E_{k} &= E_{p} && \text{they are equal} \cr E_{k} &= m g \Delta h \cr 2400 &= 10.03 \times 9.8 \times \Delta h && \text{substitute values} \cr 2400 &= 98.294h \cr \Delta h &= \frac{2400}{98.294} \cr \Delta h &= 24.41m \end{aligned} \]
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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.
We can relate the displacement of a spring to its spring constant and the force required to create the displacement using Hooke’s Law.
\[ \begin{aligned} F &= -kx && \text{Force exerted by spring} \cr \end{aligned} \]
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.
\[ \begin{aligned} & k = \frac{F}{x} \cr & k = \frac{55 \times 9.8}{0.003} \cr & k = 179667Nm^{-1} \cr & k = 1.8\times10^{5}Nm^{-1} \end{aligned} \]
\[ \begin{aligned} & E_{p} = \frac{1}{2} k x^{2} \cr & E_{p} = \frac{1}{2} \times 1.8\times10^{5} \times 0.003^{2} \cr & E_{p} = 0.81J \end{aligned} \]
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 \(80.0cm\).
\[ \begin{aligned} & k = \frac{F}{x} \cr & k = \frac{0.5 \times 9.8}{0.32} \cr & k = 15.31Nm^{-1} \end{aligned} \]
\[ \begin{aligned} & x = \frac{F}{k} \cr & x = \frac{0.9 \times 9.8}{15.31} \cr & x = 0.576m \end{aligned} \]
\[ \begin{aligned} & E_{p} = \frac{1}{2}kx^{2} \cr & E_{p} = \frac{1}{2} \times 15.31 \times 0.576^{2} \cr & E_{p} = 2.54J \end{aligned} \]
\[ \begin{aligned} & x = \frac{F}{k} \cr & x = \frac{0.5 \times 9.8}{30.62} \cr & x = 0.16m \end{aligned} \]
It halves the amount that the spring extends, and reduces the amount of energy stored by a lot.
\[ \begin{aligned} & E_{p} = \frac{1}{2}kx^{2} \cr & E_{p} = \frac{1}{2} \times 30.62 \times 0.16^{2} \cr & E_{p} = 0.39J \end{aligned} \]
\[ \begin{aligned} & E_{k} = \frac{1}{2}mv^{2} && \text{kinetic energy} \cr & E_{p} = m g \Delta h && \text{gravitational potential energy} \cr & E_{p} = \frac{1}{2} k x^{2} && \text{elastic potential energy} \end{aligned} \]
\[ \begin{aligned} & F = kx && \text{Hooke's Law} \end{aligned} \]
Step 1: Weight per Spring
\[ \begin{aligned} & F = \frac{357 \times 9.8}{4} \cr & F = 874.65N \end{aligned} \]
Step 2: Displacement
\[ \begin{aligned} & F = kx && \text{Hooke's Law} \cr & x = \frac{F}{k} \cr & x = \frac{874.65}{2.26 \times 10^{4}} \cr & x = 0.0387m \cr \end{aligned} \]
\[ \begin{aligned} & E_{p} = \frac{1}{2}kx^{2} \cr & E_{p} = \frac{1}{2} 2.26 \times 10^{4} \times 0.12^{2} \cr & E_{p} = 160J \end{aligned} \]
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The amount of energy transferred/transformed (Joules, J).
\[ \begin{aligned} & W = Fd \cr & work = force \times distance \end{aligned} \]
\[ \begin{aligned} W &= F_{w}d \cr W &= (m \times g) d \cr W &= mgh \end{aligned} \]
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?
\[ \begin{aligned} & m=500kg, d=h=1.25m && \text{(K)} \cr & W=? && \text{(U)} \cr & W=Fd=mgh && \text{(F)} \cr & W=500\times9.81\times1.25= 6131.25J && \text{(S+S)} \end{aligned} \]
The rate at which energy is transferred/transformed (the rate at which work is done).
\[ \begin{aligned} & P = \frac{W}{t} \cr & power = \frac{work}{time} \cr & power = \frac{Joules}{seconds} \cr & power = Js^{-1} && \text{also known as a Watt (W)} \end{aligned} \]
If it took Eddie \(7s\) to do \(6125J\) of work on the weights, what power was he exerting?
\[ \begin{aligned} & W = 6125J, t=7s && \text{(K)} \cr & P = ? && \text{(U)} \cr & P = \frac{W}{t} && \text{(F)} \cr & P = \frac{6125}{7} = 875Js^{-1} && \text{(S+S)} \end{aligned} \]