Newton's second law relating resultant force, mass and acceleration, the effect of friction, and calculating work done and power.

What this key area is asking

The SQA wants you to apply Newton's second law ($F = ma$) to find the resultant force, mass or acceleration of a body, to account for the effect of friction opposing motion, and to calculate the work done by a force and the power developed. These dynamics ideas describe how mechanisms and machines actually move loads.

Newton's second law

The crucial word is resultant: $F$ is the single net force left after combining all the forces acting. If the driving force is partly cancelled by friction, you use the difference. Rearrange as $a = F/m$ to find acceleration, or $m = F/a$ to find mass.

Friction

Friction is a force that always acts to oppose the relative motion (or tendency to move) between surfaces in contact. It must be subtracted from the driving force when finding the resultant, so a 30 N pull against 6 N of friction gives a 24 N resultant. Friction converts useful kinetic energy into heat, which is why it lowers the efficiency of every real machine and why moving parts get warm. It is not always unwanted: friction is what lets tyres grip and brakes work.

Work and power

Work is done when a force moves its point of application: lifting a load, or pushing a trolley along. Power is how fast that work is done. For steady motion the two power forms are equivalent, since $P = W/t = Fd/t = Fv$. Doing the same work in less time needs more power.

Putting it together

Dynamics ties force to motion and to energy. A resultant force accelerates a mass ($F = ma$); the same force doing work over a distance transfers energy ($W = Fd$); and the rate of that transfer is power ($P = Fv$). Friction sits across all three, opposing the motion, reducing the resultant force, and turning useful energy into heat. This is the link back to efficiency in the contexts area: the friction in a mechanism is exactly the energy that does not reach the output.

Examples in context

A lift raising a load shows work and power directly: the work is the weight times the height, and the motor power is that work divided by the time, or the weight times the steady raising speed. A conveyor pushing boxes against friction needs a driving force that overcomes the friction; only the resultant accelerates the boxes up to belt speed. A car braking is friction doing negative work, converting the car's kinetic energy into heat in the brakes. Recognising which force is resultant, and which energy is useful versus lost to friction, is what these calculations test.

Try this

Q1. A resultant force of 20 N acts on a 4 kg mass. Find the acceleration. [2 marks]

• Cue. $a = \frac{F}{m} = \frac{20}{4} = 5\ \text{m/s}^2$.

Q2. A force of 60 N moves a box 5 m in the direction of the force. Find the work done. [1 mark]

• Cue. $W = Fd = 60 \times 5 = 300\ \text{J}$.

Q3. A motor raises a load at 0.5 m/s with a force of 800 N. Find the power. [2 marks]

• Cue. $P = Fv = 800 \times 0.5 = 400\ \text{W}$.