1. **State the problem:** We have a meter stick suspended at its center (the 0.5 m mark), with a 1 kg object placed at $x=0$. We need to find the position $x$ where a 2 kg object should be placed to balance the stick. 2. **Set the pivot and forces:** The pivot is at the center of the meter stick, at $0.5$ m. 3. **Calculate torque due to 1 kg object:** Torque is force times distance from pivot. The distance of the 1 kg object from the pivot is $0.5 - 0 = 0.5$ m. Torque from 1 kg object: $1\times0.5 = 0.5$ (clockwise). 4. **Place the 2 kg object so that its torque balances the 1 kg torque:** The 2 kg should create an equal torque in the opposite (counterclockwise) direction. Let the position where 2 kg object is placed be $x$. Distance from pivot is $|x - 0.5|$. Set torque equality: $$2 \times |x - 0.5| = 0.5$$ 5. **Solve for $x$:** $$|x - 0.5| = \frac{0.5}{2} = 0.25$$ So, $$x - 0.5 = \pm 0.25$$ This gives two possible positions: $$x = 0.75 \quad \text{or} \quad x = 0.25$$ 6. **Interpretation:** - At $x=0.75$ m, the 2 kg object is right side of the pivot. - At $x=0.25$ m, the 2 kg object is left side of the pivot. Since 1 kg is at the left end ($x=0$), the heavier 2 kg object must be placed on the opposite side to balance. Therefore, the 2 kg object should be placed at $x=0.75$ m (right side of the center). **Final answer:** The 2 kg object should be placed at $x=0.75$ meters to balance the meter stick.