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Algebra, STD2 EQ-Bank 07 MC

The distance (\(d\)) a spring stretches varies directly with the force (\(F\)) applied to it.

When a force of 18 newtons is applied, the spring stretches 27 mm.

What is the value of the constant of variation (\(k\))?

  1. \(\dfrac{2}{5}\)
  2. \(\dfrac{5}{2}\)
  3. \(\dfrac{2}{3}\)
  4. \(\dfrac{3}{2}\)
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\(D\)

Show Worked Solution

\(d \propto F\)

\(d=kF\)

\(\text{When } d = 27 \text{ and } F = 18:\)

\(27\) \(=k \times 18\)
\(k\) \(=\dfrac{27}{18}\)
\(k\) \(=\dfrac{3}{2}\)

  
\(\Rightarrow D\)