1. **State the problem:** A spring is loaded with a 5.0 kg mass, causing its length to increase from 20 cm to 36 cm. We need to find: a) The spring constant $k$. b) The new length of the spring when a 2.0 kg mass is applied. 2. **Relevant formulas and rules:** - The force due to gravity on a mass $m$ is $F = mg$, where $g = 9.8$ m/s$^2$. - Hooke's Law for springs states $F = kx$, where $k$ is the spring constant and $x$ is the extension. - The extension $x$ is the change in length: $x = L - L_i$. 3. **Part (a) Find the spring constant $k$:** - Given mass $m = 5.0$ kg, so force $F = 5.0 \times 9.8 = 49$ N. - Initial length $L_i = 20$ cm = 0.20 m. - Final length $L = 36$ cm = 0.36 m. - Extension $x = L - L_i = 0.36 - 0.20 = 0.16$ m. - Using Hooke's Law: $$k = \frac{F}{x} = \frac{49}{0.16}$$ - Show cancellation step: $$k = \frac{49}{\cancel{0.16}} = 306.25 \text{ N/m}$$ (Note: The user calculated $k=122.5$ N/m using $x=0.4$ m, but the extension is $0.16$ m, so we correct it here.) 4. **Part (b) Find the new length with a 2.0 kg load:** - Force $F = 2.0 \times 9.8 = 19.6$ N. - Extension $x = \frac{F}{k} = \frac{19.6}{306.25}$ - Show cancellation step: $$x = \frac{19.6}{\cancel{306.25}} = 0.064 \text{ m}$$ - New length: $$L = L_i + x = 0.20 + 0.064 = 0.264 \text{ m} = 26.4 \text{ cm}$$ **Final answers:** - a) Spring constant $k = 306.25$ N/m - b) New length $L = 26.4$ cm