The marking pattern never changes
List the quantities and convert. Write the equation. Substitute. Unit. Sense-check. Use $g=10\,\text{N/kg}$ unless the paper prints another value. Square $v$ before you multiply by $\tfrac{1}{2}m$.
4.11–4.12 · Work done
$W = F \times d$ along the force
$$W = F \times d$$Work done is the energy transferred when a force moves an object in the direction of that force. $W$ is in joules, $F$ in newtons, $d$ in metres. Specification 4.12 is the link: work done = energy transferred. If you do $50\,\text{J}$ of work lifting a box, the gravitational store increases by $50\,\text{J}$.
Board example
A student pushes a crate with a force of $40\,\text{N}$ for $3.0\,\text{m}$ along the floor.
$F=40\,\text{N}$, $d=3.0\,\text{m}$.
$$W = 40 \times 3.0 = 120\,\text{J}$$That $120\,\text{J}$ goes to the thermal store of the floor and crate (friction) if the crate’s speed is constant. If the crate speeds up, some of it also fills the kinetic store.
Distance must be in the force direction
Carrying a bag horizontally at constant height does no work against gravity — the weight is vertical, the displacement is horizontal. Lifting the bag $0.80\,\text{m}$ does $W=mgh$. If a force is at an angle, this course almost always gives you the component along the motion, or asks you to use the distance moved in the direction of the force.
4.13 · Gravitational store
$\Delta GPE = m g h$
$$\Delta GPE = m g h$$$m$ in kg, $g=10\,\text{N/kg}$, $h$ in metres (the vertical height change). This is the change in the gravitational store when you lift or lower something. It is also the work you do against gravity: $W=F d$ with $F=mg$ and $d=h$.
Board example — brick
A brick of mass $2.0\,\text{kg}$ is lifted $5.0\,\text{m}$. $g=10\,\text{N/kg}$.
$$\Delta GPE = 2.0 \times 10 \times 5.0 = 100\,\text{J}$$Sense-check: a bag of sugar lifted onto a high shelf is tens of joules, not kilojoules. A person climbing a flight of stairs is thousands of joules.
Height, not path length
A ramp of length $8\,\text{m}$ that rises $2\,\text{m}$ uses $h=2\,\text{m}$ in $mgh$. The $8\,\text{m}$ is the distance you push, used in $W=Fd$ if they give the pushing force.
4.14 · Kinetic store
$KE = \tfrac{1}{2} m v^2$
$$KE = \tfrac{1}{2} m v^2$$Square the speed first. Then multiply by the mass, then by $\tfrac{1}{2}$. $v$ must be in m/s. If they give km/h, divide by $3.6$.
Board example — no drag
The $2.0\,\text{kg}$ brick falls $5.0\,\text{m}$ from rest with no air resistance. $GPE$ lost $=100\,\text{J}$, so $KE$ gained $=100\,\text{J}$.
$$v^2 = \frac{2 \times KE}{m} = \frac{2 \times 100}{2.0} = 100 \qquad v = 10\,\text{m/s}$$The classic wrong answer
$\tfrac{1}{2} \times 2.0 \times 10 = 10$, then “squared somehow” — no. $v^2 = 10^2 = 100$ first. Doubling the speed quadruples the kinetic store, because of the square.
4.15 · Conservation with and without friction
Where the joules go when something falls
A falling object empties its gravitational store. Those joules have to go somewhere.
No drag / no friction
$$mgh = \tfrac{1}{2} m v^2$$All of the gravitational store becomes kinetic. The $m$ cancels if you are finding $v$: $v=\sqrt{2gh}$.
With friction or drag
$$mgh = KE + W_{\text{against friction}}$$Work is done against friction, so the thermal store of the object and surroundings increases. $KE$ at the bottom is smaller than $mgh$.
Board example — friction takes $20\,\text{J}$
Same brick: $mgh=100\,\text{J}$. Work done against friction $=20\,\text{J}$.
$$KE = 100 - 20 = 80\,\text{J}$$The $20\,\text{J}$ is now in the thermal store. The speed is less than $10\,\text{m/s}$.
A roller-coaster or skateboard down a slope is the same bookkeeping. If they give you $KE$ at the bottom and $mgh$ at the top, the difference is work against friction.
3-mark conservation explain
Gravitational store decreases by $mgh$ [1]. Without friction this all transfers to the kinetic store [1]. With friction, some energy is transferred mechanically to the thermal store of the surroundings, so $KE < mgh$ [1].
4.16–4.17 · Power
Rate of doing work, or of transferring energy
$$P = \frac{W}{t} = \frac{E}{t}$$Power is how fast energy is transferred. $P$ in watts, $W$ or $E$ in joules, $t$ in seconds. A $100\,\text{W}$ lamp transfers $100\,\text{J}$ every second. Two students can do the same work lifting a box; the one who does it in less time has a larger power.
Board example
A motor does $2400\,\text{J}$ of work in $8.0\,\text{s}$.
$$P = \frac{2400}{8.0} = 300\,\text{W}$$You can also write $P=E/t$ or, if they give a force and a constant speed, $P=F v$ because $W=Fd$ and $d=vt$.
Joules are not watts
Energy is a pile of joules. Power is joules per second. A $2\,\text{kW}$ kettle running for $30\,\text{s}$ transfers $E=Pt=2000\times 30=60\,000\,\text{J}$. Mixing $E$ and $P$ is the most common unit error on this paper.
Watch
Kinetic energy, gravitational energy and power
Which equation?
| They give you… | They want… | Use |
|---|---|---|
| Force and distance along the force | energy transferred | $W=Fd$ |
| Mass, $g$, vertical height | change in gravitational store | $mgh$ |
| Mass and speed | kinetic store | $\tfrac{1}{2}mv^2$ |
| A fall, “no air resistance” | speed at the bottom | $mgh=\tfrac{1}{2}mv^2$ |
| A fall plus friction / heat | $KE$ or work against friction | $mgh = KE + W_f$ |
| Energy (or work) and time | power | $P=W/t$ |
Check you can
1.
A force of $25\,\text{N}$ pushes a trolley $4.0\,\text{m}$. Calculate the work done.
$W=Fd=25\times 4.0=100\,\text{J}$.
2.
A $50\,\text{kg}$ skateboarder drops $3.2\,\text{m}$. $g=10\,\text{N/kg}$. Friction does $400\,\text{J}$ of work. Find $GPE$, $KE$ at the bottom, and $v$.
$GPE=50\times 10\times 3.2=1600\,\text{J}$. $KE=1600-400=1200\,\text{J}$. $v=\sqrt{2\times 1200/50}=\sqrt{48}\approx 6.9\,\text{m/s}$.
3.
A student does $1800\,\text{J}$ of work in $12\,\text{s}$. Calculate her power.
$P=W/t=1800/12=150\,\text{W}$.