The spring stretches 20 cm (0.2 m) under a mass of 5 kg. The force due to gravity is \( F = mg = 5 \times 9.8 = 49 \, \text{N} \).
Using Hooke's Law, \( F = kx \), we have:
\[ k \times 0.2 = 49 \]
\[ k = \frac{49}{0.2} = 245 \, \text{N/m} \]
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