Newton's laws of motion, free-body diagrams and resolving forces, balanced and unbalanced forces, motion on a slope, tension, and work done, kinetic and potential energy and power.

What this key area is asking

The SQA wants you to apply Newton's three laws, draw free-body diagrams and resolve forces (especially on a slope), distinguish balanced from unbalanced forces, handle tension, and use the work-energy ideas: work done, kinetic energy, gravitational potential energy and power.

Newton's laws of motion

Balanced forces (zero net force) mean zero acceleration, so the object is either stationary or moving at constant velocity. Unbalanced forces mean acceleration. Recognising "constant velocity" as "balanced forces" is the key to many slope and tension problems.

Free-body diagrams and resolving on a slope

A free-body diagram shows every force acting on a single object as an arrow. To handle a slope, resolve the weight $mg$ into a component along the slope and a component perpendicular to it.

The component down the slope is what tends to accelerate an object downhill; the perpendicular component is balanced by the normal reaction. If the object moves at constant velocity up a frictionless slope, the applied force or tension exactly equals $mg\sin\theta$.

Work, energy and power

Work done is the energy transferred by a force. By the work-energy principle, the net work done on an object equals its change in kinetic energy. Power is the rate of transferring energy; on a vehicle moving at steady speed $v$ against a resistive force $F$, the engine power is $P = Fv$.

Examples in context

A funicular railway or a ski lift must supply a tension equal to the weight component $mg\sin\theta$ of the cars along the slope, plus extra to accelerate; engineers size the motor from $P = Fv$. A lift (elevator) accelerating upward needs a cable tension greater than the weight ($T - mg = ma$), illustrating Newton's second law. A rocket rises because the downward thrust on the exhaust gases produces an equal and opposite upward force on the rocket (third law). A cyclist at steady speed develops power $P = Fv$ to overcome air resistance and friction; doubling the speed against air drag needs far more power. Roller coasters convert gravitational potential energy $mgh$ at the top of a drop into kinetic energy $\tfrac{1}{2}mv^2$ at the bottom.

Try this

Q1. State Newton's second law as an equation. [1 mark]

• Cue. $F = ma$ (unbalanced force equals mass times acceleration).

Q2. A $3.0\ \text{kg}$ object rests on a slope at $30^{\circ}$. Calculate the component of its weight acting down the slope. Take $g = 9.8\ \text{N kg}^{-1}$. [2 marks]

• Cue. $F_{\parallel} = mg\sin\theta = 3.0 \times 9.8 \times \sin 30^{\circ} = 14.7\ \text{N}$.

Q3. A motor lifts a $50\ \text{kg}$ load a height of $4.0\ \text{m}$ in $8.0\ \text{s}$. Calculate the power. Take $g = 9.8\ \text{N kg}^{-1}$. [3 marks]

• Cue. $E_p = mgh = 50 \times 9.8 \times 4.0 = 1960\ \text{J}$; $P = \frac{E}{t} = \frac{1960}{8.0} = 245\ \text{W}$.