Physics 9702 · Paper 2

Work at an Angle

Forces are often not lined up with the motion: a rope pulls upward at an angle while the box slides along the floor.
Only the part of the force along the motion does work. One cosine handles it.

Only the along-the-motion part counts

W = Fs\cos\theta

Symbols

$W$ = work done (J)
$F$ = force (N)
$s$ = displacement (m)
$\theta$ = angle between the force and the displacement (no unit (degrees or radians))

The box moves horizontally, so only the horizontal part of the pull, F cos 30°, does work: W = F s cos 30°. The vertical part does no work on the box.

$F\cos\theta$ is the component of the force along the motion: the resolving skill from lesson 1.9.
Three special cases follow at once. Force along the motion: $\cos 0^\circ = 1$ , full work. Force at 90°: $\cos 90^\circ = 0$ , zero work. Force against the motion: $\cos 180^\circ = -1$ , negative work (the force takes energy away).
Friction and drag point against the motion, so they do negative work: a 12 N friction force over 3.0 m takes away $12 \times 3.0 = 36 \text{ J}$ .

Worked example

One change: gravity on a slope

A stone of weight 5.0 N rolls 50 m down a slope, dropping a vertical height of 30 m. Find the work done by gravity.

Gravity acts vertically, so only the vertical part of the movement counts: 30 m, not 50 m.
$W = 5.0 \times 30 = 150 \text{ J}$ .

Answer

W = 150 \text{ J}

Worked example

Exam version: pulled by a kite wire

A skier moves at a constant 4.4 m s⁻¹ for 30 s, pulled by a wire at 30° above the horizontal with tension 140 N. Find the work done by the tension. (9702/22/F/M/20 Q3(b)(i))

Displacement first: $s = vt = 4.4 \times 30 = 132 \text{ m}$ .
Only the horizontal part of the tension does work: $W = 140\cos 30^\circ \times 132 = 1.6 \times 10^{4} \text{ J}$ .

Answer

W = 1.6 \times 10^{4} \text{ J}

Common mistake

Multiplying the full force by the full path length without checking directions. First ask: how far did the object move along this force's direction? A weight moved horizontally p and vertically q does work

Wq

, not

W \times

the diagonal (9702/13/O/N/24 Q19).

A useful zero: throw a ball up to height h and let it come back to the same height. Gravity did negative work going up and equal positive work coming down: total zero (9702/11/O/N/24 Q18).

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Work at an Angle

Forces are often not lined up with the motion: a rope pulls upward at an angle while the box slides along the floor.
Only the part of the force along the motion does work. One cosine handles it.

Only the along-the-motion part counts

W = Fs\cos\theta

Symbols

$W$ = work done (J)
$F$ = force (N)
$s$ = displacement (m)
$\theta$ = angle between the force and the displacement (no unit (degrees or radians))

The box moves horizontally, so only the horizontal part of the pull, F cos 30°, does work: W = F s cos 30°. The vertical part does no work on the box.

$F\cos\theta$ is the component of the force along the motion: the resolving skill from lesson 1.9.
Three special cases follow at once. Force along the motion: $\cos 0^\circ = 1$ , full work. Force at 90°: $\cos 90^\circ = 0$ , zero work. Force against the motion: $\cos 180^\circ = -1$ , negative work (the force takes energy away).
Friction and drag point against the motion, so they do negative work: a 12 N friction force over 3.0 m takes away $12 \times 3.0 = 36 \text{ J}$ .

Worked example

One change: gravity on a slope

A stone of weight 5.0 N rolls 50 m down a slope, dropping a vertical height of 30 m. Find the work done by gravity.

Gravity acts vertically, so only the vertical part of the movement counts: 30 m, not 50 m.
$W = 5.0 \times 30 = 150 \text{ J}$ .

Answer

W = 150 \text{ J}

Worked example

Exam version: pulled by a kite wire

A skier moves at a constant 4.4 m s⁻¹ for 30 s, pulled by a wire at 30° above the horizontal with tension 140 N. Find the work done by the tension. (9702/22/F/M/20 Q3(b)(i))

Displacement first: $s = vt = 4.4 \times 30 = 132 \text{ m}$ .
Only the horizontal part of the tension does work: $W = 140\cos 30^\circ \times 132 = 1.6 \times 10^{4} \text{ J}$ .

Answer

W = 1.6 \times 10^{4} \text{ J}

Common mistake

Wq

, not

W \times

the diagonal (9702/13/O/N/24 Q19).

A useful zero: throw a ball up to height h and let it come back to the same height. Gravity did negative work going up and equal positive work coming down: total zero (9702/11/O/N/24 Q18).