Apply: energy transferred in stretching (J) = ½ × spring constant (N/m) ×…

The idea

Stretching a spring stores elastic potential energy. The energy transferred is E = ½ × k × e², where E is energy in joules (J), k is the spring constant in N/m and e is the extension in metres. The extension is squared, so it must be in metres before squaring; 4 cm becomes 0.04 m, and 0.04² = 0.0016 m². A spring with k = 300 N/m stretched by 0.02 m stores E = ½ × 300 × 0.0004 = 0.06 J. Because of the square, doubling the extension quadruples the energy stored. The formula applies within the elastic limit, where the spring returns to its original shape.

What earns the marks

• Selects the equation E = ½ × k × e² and substitutes the spring constant in N/m and the extension in m.
• Squares the extension after converting it to metres, for example 0.04² = 0.0016 m².
• Multiplies by one half as well as by the spring constant and the squared extension.
• Gives the energy in joules with the correct unit label.
• Recognises that doubling the extension increases the stored energy by a factor of four.

Common mistakes and how to put them right

• Squares the extension in centimetres instead of converting to metres first.

  Put it right: Convert then square.

• Forgets the factor of one half.

  Put it right: Include ½ in the calculation.

• Multiplies extension by two instead of squaring it; use e × e.

• Reports energy in newtons or N/m.

  Put it right: The correct unit is the joule (J).

Exam tips

• Convert extension to metres before squaring, then substitute into E = ½ × k × e².
• Write out the squared value separately so the arithmetic is easy to check.
• Check the magnitude: energy rises with the square of extension, so small changes in e matter a lot.