1. **State the problem:** We have a spring with a hanger whose position changes linearly with the mass applied. The hanger is 30.6 cm off the ground with no mass, 16.2 cm off the ground with 150 g, and 23.3 cm off the ground with an unknown mass (wooden block). We need to find the mass of the wooden block. 2. **Set up the linear relationship:** Let $m$ be the mass in grams and $y$ be the hanger position in cm. The relationship is linear: $$y = mx + b$$ where $x$ is the mass and $b$ is the position with zero mass. 3. **Identify known points:** - At $x=0$, $y=30.6$ cm (no mass) - At $x=150$, $y=16.2$ cm 4. **Find the slope $m$:** $$m = \frac{y_2 - y_1}{x_2 - x_1} = \frac{16.2 - 30.6}{150 - 0} = \frac{-14.4}{150} = -0.096$$ 5. **Write the equation of the line:** $$y = -0.096x + 30.6$$ 6. **Use the position for the wooden block to find its mass:** Given $y=23.3$, solve for $x$: $$23.3 = -0.096x + 30.6$$ 7. **Isolate $x$:** $$23.3 - 30.6 = -0.096x$$ $$-7.3 = -0.096x$$ 8. **Divide both sides by $-0.096$:** $$x = \frac{-7.3}{-0.096}$$ $$x = \frac{\cancel{-7.3}}{\cancel{-0.096}} = 76.0$$ 9. **Final answer:** The mass of the wooden block is approximately **76.0 grams**.