Resultant Force & Acceleration | CAIE 9702 Paper 1 Notes

Chapters

Paper 1

Physics9702

Resultant Force & Acceleration

Add every force with direction before using F = ma.

If a force is angled, resolve only the component along the acceleration.

Find the resultant force first

Most exam questions give more than one force, so the working always has the same first move: add the forces with signs.

Worked example

One change: two opposite forces

A car of mass 500 kg has a forward driving force of 300 N and air resistance of 200 N. Find the acceleration.

Resultant force, taking forward as positive: $F = 300 - 200 = 100 \text{ N}$ .
$a = \dfrac{F}{m} = \dfrac{100}{500} = 0.20 \text{ m s}^{-2}$ .

Answer

a = 0.20 \text{ m s}^{-2}

Worked example

One change: slowing down (negative answer)

A car of mass 500 kg travels at 20 m s⁻¹. The driver brakes and the car stops in 10 s. Find the braking force.

Acceleration first: $a = \dfrac{v-u}{t} = \dfrac{0-20}{10} = -2.0 \text{ m s}^{-2}$ .
$F = ma = 500 \times (-2.0) = -1000 \text{ N}$ .
The minus sign means the force points against the motion. The brakes provide 1000 N backwards.

Answer

F = 1000 \text{ N opposite to the motion}

Common mistake

Putting a single force into

F = ma

when the object has several forces on it.

F

must be the resultant. If a question gives a driving force and a resistance, subtract before you divide by the mass.

The exam version: a force at an angle

In Paper 2 the pull often acts at an angle, along a rope or wire, so one resolving step comes first.

Only the horizontal part of the pull (T cos 28° = 212 N) drives the skier forward; the vertical part (T sin 28° = 113 N) just lifts a little. Split the pull first, then use 212 − R = ma to get the resistance.

Worked example

Kite-skier: resolve, then F = ma

A skier of mass 89 kg is pulled over level snow by a wire at 28° to the horizontal. The tension in the wire is 240 N and the skier accelerates at 0.80 m s⁻². Find the total resistive force R on the skier. (9702/23/M/J/20 Q2(b)(iii))

Horizontal part of the tension: $240\cos 28^\circ = 212 \text{ N}$ .
Resultant force needed for the acceleration: $F = ma = 89 \times 0.80 = 71.2 \text{ N}$ .
The resultant is the pull minus the resistance: $71.2 = 212 - R$ .
$R = 212 - 71.2 = 141 \approx 140 \text{ N}$ .

Answer

R = 140 \text{ N}

Ways the exam changes this question: find the acceleration instead of the force; find one missing force when $a$ is given; feed the acceleration into a SUVAT equation from chapter 2 to get a speed or a distance.
Vertical versions use weight as one of the forces. Weight is the next lesson.

Your turnAdapted from 9702/23/M/J/20 Q2(b)(iii)

A 60 kg skier is pulled horizontally by 180 N while a resistive force of 60 N acts backwards. Calculate the acceleration.

Show worked answer

Resultant force is $180-60=120\ \text{N}$ .
$a=F/m=120/60=2.0\ \text{m s}^{-2}$ .

Answer

a=2.0\ \text{m s}^{-2}

Resultant Force & Acceleration

Add every force with direction before using F = ma.

If a force is angled, resolve only the component along the acceleration.

Find the resultant force first

Most exam questions give more than one force, so the working always has the same first move: add the forces with signs.

Worked example

One change: two opposite forces

A car of mass 500 kg has a forward driving force of 300 N and air resistance of 200 N. Find the acceleration.

Resultant force, taking forward as positive: $F = 300 - 200 = 100 \text{ N}$ .
$a = \dfrac{F}{m} = \dfrac{100}{500} = 0.20 \text{ m s}^{-2}$ .

Answer

a = 0.20 \text{ m s}^{-2}

Worked example

One change: slowing down (negative answer)

A car of mass 500 kg travels at 20 m s⁻¹. The driver brakes and the car stops in 10 s. Find the braking force.

Acceleration first: $a = \dfrac{v-u}{t} = \dfrac{0-20}{10} = -2.0 \text{ m s}^{-2}$ .
$F = ma = 500 \times (-2.0) = -1000 \text{ N}$ .
The minus sign means the force points against the motion. The brakes provide 1000 N backwards.

Answer

F = 1000 \text{ N opposite to the motion}

Common mistake

Putting a single force into

F = ma

when the object has several forces on it.

F

must be the resultant. If a question gives a driving force and a resistance, subtract before you divide by the mass.

The exam version: a force at an angle

In Paper 2 the pull often acts at an angle, along a rope or wire, so one resolving step comes first.

Worked example

Kite-skier: resolve, then F = ma

Horizontal part of the tension: $240\cos 28^\circ = 212 \text{ N}$ .
Resultant force needed for the acceleration: $F = ma = 89 \times 0.80 = 71.2 \text{ N}$ .
The resultant is the pull minus the resistance: $71.2 = 212 - R$ .
$R = 212 - 71.2 = 141 \approx 140 \text{ N}$ .

Answer

R = 140 \text{ N}

Ways the exam changes this question: find the acceleration instead of the force; find one missing force when $a$ is given; feed the acceleration into a SUVAT equation from chapter 2 to get a speed or a distance.
Vertical versions use weight as one of the forces. Weight is the next lesson.

Your turnAdapted from 9702/23/M/J/20 Q2(b)(iii)

A 60 kg skier is pulled horizontally by 180 N while a resistive force of 60 N acts backwards. Calculate the acceleration.

Show worked answer

Resultant force is $180-60=120\ \text{N}$ .
$a=F/m=120/60=2.0\ \text{m s}^{-2}$ .

Answer

a=2.0\ \text{m s}^{-2}