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Physics graph skills

Read scales, plot a point and use gradients and intercepts to interpret Physics graphs.

Practise

Find a spring constant

This spring obeys F = kx over the displayed range. Use the extension–force graph to find k in N m⁻¹. Check which quantity is on each axis. Give your answer to 3 significant figures.

Extension against forceAn ideal straight line for a spring within its proportional region. Horizontal axis: Force / N, 0 to 8. Vertical axis: Extension / cm, 0 to 32. Exact values are also listed below the graph.0246808162432Force / NExtension / cmAB
Read the axes and units. The values are also available in the table below.
Larger graph
Extension against forceAn ideal straight line for a spring within its proportional region. Horizontal axis: Force / N, 0 to 8. Vertical axis: Extension / cm, 0 to 32. Exact values are also listed below the graph.0246808162432Force / NExtension / cmAB
Graph values
Coordinates
PointForce / NExtension / cm
A14
B728

Treat the displayed values as exact unless an uncertainty is stated.

Use decimals or e notation, e.g. 4.2e-3. Enter the unit separately.

Hint

Here the gradient is extension / force = 1 / k. Convert cm per N to m per N, then take the reciprocal.

Worked solution
  1. Gradient = (28 − 4) / (7 − 1) = 4 cm N⁻¹.
  2. In SI units the gradient is 0.04 m N⁻¹.
  3. k = 1 / gradient = 100 / 4 N m⁻¹.
  4. Answer: 25.0 N m⁻¹ (3 significant figures).
Another question
Question type: Mixed practice
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Calculate a graph gradient from two points

A potential-difference–current line passes through (0.10 A, 1.5 V) and (0.50 A, 7.5 V). Gradient = (7.5 − 1.5)/(0.50 − 0.10) = 15 V A⁻¹ = 15 Ω. Choose separated points on the fitted line; they need not be measured data points.

Read which quantity is on each axis. If current is vertical and potential difference horizontal, the gradient is conductance, 1/R. Dividing one coordinate by the other gives the gradient only when the relevant straight line passes through the origin.

Find uncertainty in a gradient

Read the scale before calculating

Count the small divisions between two labelled values. If 10 divisions span 2 seconds, each division is 0.2 s. Check whether the scale starts at zero.

Gradient and intercept

For a straight line, gradient = Δy / Δx. Use two well-separated points on the line and subtract in the same order. A falling line has a negative gradient. Use changes in the coordinates; y / x only gives the gradient when the line passes through the origin.

The vertical intercept is the value of y when x = 0. On a displacement–time graph it gives initial displacement, while the gradient gives velocity in m s⁻¹.

Best-fit lines

Follow the overall trend and balance points above and below the line. A best-fit line need not pass through a data point. The readings here are synthetic practice data; the exercise asks you to choose between three candidate lines.

Use the axes to interpret the gradient

For V against I, the gradient is V / I = R, in Ω. Reversing the axes gives 1 / R. For extension x against force F, x = F / k: the gradient is 1 / k. Convert extension to metres before finding k in N m⁻¹.

What this practice includes

Linear scales, plotting one point, signed gradients, intercepts, choosing a straight best-fit line, axes and two physical interpretations. Displayed values are treated as exact for practice. Numerical answers use 3 significant figures; plotting coordinates must match the stated values exactly.

Curve fitting, tangents, areas, logarithmic axes and gradient uncertainty are not included yet.

A-level relevance

AQA 7408: MS 3.2–3.5 (partial); AQA 7408: 3.4.1.3, 3.4.2.1 (selected); OCR A H556: 1.1.3(d)(i–ii), 3.1.1(c), 3.4.1(b–d) (partial). Tasks cover the selected skills listed above.

AQA specification · AQA mechanics and materials specification · OCR A specification