Energy and Kinetic Energy

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.

Unit: Work and Energy · Updated

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.

Test yourself

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