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Kinetic Energy

Mahi Tuatahi 🔗

Ngā Whāinga Ako 🔗

  1. Give the symbols and units for kinetic energy
  2. $E_{k} = \frac{1}{2}mv^{2}$

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.

$$ \begin{aligned} E_{k} &= \frac{1}{2} \times mass \times \text{velocity squared} \cr E_{k} &= \frac{1}{2} \times m \times v^{2} \end{aligned} $$

What does $v^{2}$ mean? 🔗

$$ \begin{aligned} E_{k} &= \frac{1}{2} \times m \times v \times v \end{aligned} $$

Ngā Pātai 🔗

  1. Mr LeSueur rides his bike to work at $32km/h$ ($8.89ms^{-1}$). Both he and his bike have a combined mass of $78kg$. Calculate his kinetic energy.
  2. A ball is bowled with energy $39.2J$ and velocity $28ms^{-1}$. Calculate its mass.
  3. A dog with mass $22kg$ sprints across a field with a kinetic energy of $1100J$. Calculate it’s velocity.

$$ \begin{aligned} K:& \cr U:& \cr F:& \cr S+S:& \end{aligned} $$

Whakatika Tahi 🔗

Mr LeSueur rides his bike to work at $32km/h$ ($8.89ms^{-1}$). Both he and his bike have a combined mass of $78kg$. Calculate his kinetic energy.

$$ \begin{aligned} m&=78kg, v=8.89ms^{-1} && \text{(K)} \cr E_{k} &= ? && \text{(U)} \cr E_{k} &=\frac{1}{2}mv^{2} && \text{(F)} \cr E_{k} &= \frac{1}{2} \times 78 \times 8.89^{2} && \text{(S)} \cr &= 3082.3J && \text{(S)} \end{aligned} $$

Whakatika Rua 🔗

A ball is bowled with energy $39.2J$ and velocity $28ms^{-1}$. Calculate its mass.

$$ \begin{aligned} E_{k}&=39.2J, v=28ms^{-2} && \text{(K)} \cr m &= ? && \text{(U)} \cr E_{k} &=\frac{1}{2}mv^{2} && \text{(F)} \cr 39.2 &= \frac{1}{2} \times m \times 28^{2} && \text{(S)} \cr \sqrt{\frac{39.2 \times 2}{784}} &= m = 0.1kg && \text{(S)} \end{aligned} $$

Whakatika Toru 🔗

A dog with mass $22kg$ sprints across a field with a kinetic energy of $1100J$. Calculate it’s velocity.

$$ \begin{aligned} m&=22kg, E_{k}=1100J && \text{(K)} \cr v &= ? && \text{(U)} \cr E_{k} &=\frac{1}{2}mv^{2} && \text{(F)} \cr 1100 &= \frac{1}{2} \times 22 \times v^{2} && \text{(S)} \cr \sqrt{\frac{1100 \times 2}{22}} &= v = 10ms^{-1} && \text{(S)} \end{aligned} $$

Alternatively, you can rearrange the formula first, and then substitute in the numbers.

$$ \begin{aligned} m&=22kg, E_{k}=1100J, v=? \cr E_{k} &= \frac{1}{2}mv^{2} \cr 2E_{k} &= mv^{2} \cr \frac{2E_{k}}{m} &= v^{2} \cr \sqrt{\frac{2E_{k}}{m}} &= v = \sqrt{\frac{2 \times 1100}{22}} = 10ms^{-1} \end{aligned} $$

Whakamātau: Finding Your Kinetic Energy 🔗

Open the whakamātau document on Google Classroom!

Whakawai / Practice 🔗