Energy and Kinetic Energy: MDCAT Physics notes
Energy and Kinetic Energy notes for MDCAT: work–energy principle, KE = ½mv², KE–momentum relation p²/2m, doubling speed, spring stopping problems.
Energy
Energy is the ability of a body to do work. Work done on a body is stored in it as energy, so energy and work share the same unit, the joule. The Sun is the main direct source of energy reaching the Earth; wind, hydro and fossil fuels are ultimately derived from it.
Kinetic energy
Kinetic energy is the energy a body has due to its motion. If a force $F$ accelerates a mass $m$ from rest through distance $d$ to speed $v$, the work done is $Fd = ma \cdot \frac{v^2}{2a}$, so
$$K.E. = \tfrac12 mv^2$$
- KE is a scalar and can never be negative (mass and $v^2$ are both positive).
- Minimum KE is zero, for a body at rest.
- $K.E. \propto v^2$: doubling speed makes KE four times; halving speed makes it one quarter.
Example: a 0.5 kg ball at $6\ \text{m s}^{-1}$ has $K.E. = \tfrac12(0.5)(36) = 9\ \text{J}$.
Work–energy principle
The work done by the net force on a body equals the change in its kinetic energy:
$$W_{net} = \tfrac12 mv_f^2 - \tfrac12 mv_i^2$$
An 800 kg car speeding up from 20 to $30\ \text{m s}^{-1}$: $\Delta K.E. = \tfrac12(800)(900-400) = 2\times10^5\ \text{J} = 200\ \text{kJ}$.
Stopping by a spring: an 8.0 kg box at $3\ \text{m s}^{-1}$ has $K.E. = 36\ \text{J}$. If it compresses a spring 0.12 m before stopping, the average retarding force is $F = K.E./d = 36/0.12 = 300\ \text{N}$.
Kinetic energy and momentum
Since $p = mv$:
$$K.E. = \frac{p^2}{2m}, \qquad p = \sqrt{2m\,K.E.}$$
- Same momentum: $K.E. \propto 1/m$. If KEs are in the ratio 4 : 1, masses are in the ratio 1 : 4.
- Same KE: $p \propto \sqrt m$. A train and a car with equal KE: the heavier train has greater momentum.
- If KE becomes 4 times (same mass), momentum becomes $\sqrt4 = 2$ times.
Force acting for a time
A force $P$ acting for time $t$ on a mass at rest gives momentum $p = Pt$, so
$$K.E. = \frac{P^2t^2}{2m}$$
KE of a projectile at the top
At the highest point only $v\cos\theta$ remains, so $K.E._{top} = E\cos^2\theta$. For $\theta = 45^\circ$, $K.E._{top} = E/2$.
Key formulas
- $K.E. = \tfrac12 mv^2 = p^2/2m$
- $W_{net} = \Delta K.E.$
- Average stopping force: $F = K.E./d$
- Projectile top: $K.E. = E\cos^2\theta$
Common MDCAT traps
- $K.E. = p^2/2m$, not $p/2m$ or $p^2/2m^2$.
- Doubling speed quadruples KE; doubling KE multiplies momentum by $\sqrt2$ only.
- KE at the top of a $45^\circ$ projectile is $E/2$, not zero.
- Equal KE: the heavier body has more momentum.
- KE is never negative; its minimum is zero.
Quick revision
- Unit of KE = unit of work = joule.
- Body at rest: KE zero.
- Halve the speed: KE becomes 1/4.
- Equal momentum: lighter body has more KE.
- Work done on a body is stored as energy.