Energy & Hooke’s Law

12PHYS - Mechanics

Finn LeSueur

2021

Akoranga 37 Mahi Tuatahi

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

  1. How long does it take for the ball to reach its highest point?
  2. 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

  1. Be able to calculate gravitational potential energy (\(E_{p} = mgh\))
  2. Be able to calculate kinetic energy (\(E_{k} = \frac{1}{2}mv^{2}\))
  3. 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 transformed or transferred.

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 ground in 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

  1. Be able to calculate the energy stored in a spring (\(E_{p} = \frac{1}{2}kx^{2}\))
  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.

  1. Calculate the spring constant of the bench. (M)
  2. Give correct SI units for the spring constant. (A)
  3. Calculate the elastic potential energy stored in the bench. (A)

Whakatika

  1. Calculate the spring constant of the bench + unit
\[\begin{aligned} & k = \frac{F}{x} \newline & k = \frac{55 \times 9.8}{0.003} \newline & k = 179667Nm^{-1} \newline & k = 1.8\times10^{5}Nm^{-1} \end{aligned}\]
  1. 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 \(80.0cm\).

  1. Calculate the spring constant. (M)
  2. Write a unit with your answer. (A)
  3. Calculate the energy stored in the spring when a second toy of mass \(400g\) is also hung along with the aeroplane. (M)
  4. 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

  1. Calculate the spring constant. (M)
\[\begin{aligned} & k = \frac{F}{x} \newline & k = \frac{0.5 \times 9.8}{0.32} \newline & k = 15.31Nm^{-1} \end{aligned}\]
  1. Calculate the energy stored in the spring when a second toy of mass \(400g\) is also hung along with the aeroplane. (M)
\[\begin{aligned} & x = \frac{F}{k} \newline & x = \frac{0.9 \times 9.8}{15.31} \newline & x = 0.576m \end{aligned}\] \[\begin{aligned} & E_{p} = \frac{1}{2}kx^{2} \newline & E_{p} = \frac{1}{2} \times 15.31 \times 0.576^{2} \newline & E_{p} = 2.54J \end{aligned}\]
  1. 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

  1. Give the equations for kinetic, gravitational potential and elastic potential energy.
  2. Give the name and formula for the law that you can use to relate force, spring constant and displacement.
  3. 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)
  4. How much energy is stored in each spring if car sinks \(0.12m\)? (A)

Whakatika

  1. Give the equations for kinetic, gravitational potential and elastic potential energy.
\[\begin{aligned} & E_{k} = \frac{1}{2}mv^{2} && \text{kinetic energy} \newline & E_{p} = m g \Delta h && \text{gravitational potential energy} \newline & E_{p} = \frac{1}{2} k x^{2} && \text{elastic potential energy} \end{aligned}\]
  1. Give the name and formula for the law that you can use to relate force, spring constant and displacement.
\[\begin{aligned} & F = kx && \text{Hooke's Law} \end{aligned}\]
  1. 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

\[\begin{aligned} & F = kx && \text{Hooke's Law} \newline & x = \frac{F}{k} \newline & x = \frac{874.65}{2.26 \times 10^{4}} \newline & x = 0.0387m \newline \end{aligned}\]
  1. 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

  1. Be able to define & calculate work done (\(W=Fd=mgh\))
  2. 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: