1.1.2a Kinetic energy calculations

Kinetic energy: the store an object fills by moving

1. Any moving object has energy in its kinetic store, such as a moving car, a thrown ball or a running athlete.
2. If the object stops, energy is transferred out of its kinetic store, often to the thermal store of the surroundings through friction or air resistance.
3. The kinetic energy of an object depends on its mass $m$ in $kg$ and its speed $v$ in $m/s$.
4. Calculate it with $E_k=\frac{1}{2}m{v}^2$.
5. Here $E_k$ is the kinetic energy in $J$, $m$ is the mass in $kg$, and $v$ is the speed in $m/s$.
6. You should be able to recall and apply this equation.

Why speed matters more than mass

1. The equation squares the speed, $v^2$, so speed has a bigger effect than mass.
2. If the mass doubles and the speed stays the same, the kinetic energy doubles.
3. If the speed doubles and the mass stays the same, the kinetic energy becomes four times bigger, because $2^2=4$.
4. So a car travelling twice as fast has four times the kinetic energy, which is why it needs far more energy transferred out of its kinetic store to stop and usually has a longer stopping distance.

Calculating kinetic energy

1. Write the equation first, then substitute the values with their units, then calculate the answer.

Using the correct units

1. Mass must be in $kg$ and speed must be in $m/s$, and the answer then comes out in $J$.
2. If the mass is given in grams, convert it first using $1000g=1kg$, so $500g=0.500kg$.
3. If a speed is given in another unit, such as $km/h$, convert it before substituting, because the wrong units give the wrong energy.

Comparing kinetic energies without a full calculation

1. You can compare kinetic energies straight from the structure of the equation.
2. Kinetic energy is directly proportional to mass, so at the same speed an object with twice the mass has twice the kinetic energy.
3. Kinetic energy is proportional to speed squared, so at the same mass an object with twice the speed has four times the kinetic energy.

1.1.2b Elastic potential energy calculations

Elastic potential energy: energy stored by stretching

1. The elastic potential energy stored in a stretched spring depends on the spring constant $k$ in $N/m$ and the extension $e$ in $m$.
2. The spring constant describes the stiffness of the spring.
3. A spring with a larger spring constant is stiffer, so it stores more energy for the same extension.

Extension: the increase in length

1. Measure the extension from the spring's original, unstretched length, not from its total stretched length.
2. Convert the extension to metres before using it, dividing a value in centimetres by $100$.

Calculating elastic potential energy

1. Calculate the stored energy with $E_e=\frac{1}{2}k{e}^2$.
2. Here $E_e$ is the elastic potential energy in $J$, $k$ is the spring constant in $N/m$, and $e$ is the extension in $m$.
3. This equation is given to you in the exam, so you apply it rather than recall it.
4. The extension is squared, so doubling the extension makes the elastic potential energy four times as large, since $\frac{1}{2}k(2e{)}^2=\frac{1}{2}k(4e^2)$.
5. The equation only holds while the spring has not passed its limit of proportionality, beyond which the extension is no longer proportional to the force.

Rearranging the equation

1. You may need to rearrange the equation, for example to find the spring constant or the extension.
2. For the spring constant, use $k=\frac{2E_e}{e^2}$.
3. For the extension, use $e=\sqrt{\frac{2E_e}{k}}$.

1.1.2c Gravitational potential energy calculations

Gravitational potential energy: energy stored by height

1. When an object is lifted above the ground, energy is transferred to its gravitational potential store, because a force is needed to lift it against gravity.
2. The higher the object is lifted, the more gravitational potential energy it gains.
3. A more massive object gains more gravitational potential energy for the same height, because gravity acts more strongly on a larger mass.

The equation you must recall

1. You must be able to recall and apply $E_p=mgh$.
2. Here $E_p$ is the gravitational potential energy in $J$, $m$ is the mass in $kg$, $g$ is the gravitational field strength in $N/kg$, and $h$ is the height in $m$.
3. In words, gravitational potential energy $=$ mass $\times$ gravitational field strength $\times$ height.
4. In any calculation the value of $g$ will be given, so you do not need to remember it.
5. The height $h$ is the vertical height gained above the ground, not the length of a slope, staircase or path taken to get there.

Rearranging the equation

1. You may also need to rearrange the equation, for example to find the mass or the height.
2. For the mass, use $m=\frac{E_p}{gh}$.
3. For the height, use $h=\frac{E_p}{mg}$.