Question

Four bodies of masses 8 kg, 2 kg, 4 kg and 2 kg are placed at the four corners A, B, C and D respectively of a square ABCD of diagonal 80 cm. The distance of the center of mass of the system from the corner A is:

• A. 30 cm
• B. 40 cm
• C. 60 cm
• D. 20 cm

Hint:

The center of mass of a system can be found by taking the weighted average of the positions of all masses involved.

Answer 1

The Correct Option is A

- Let the original force be \(F\). When the force is increased by \(80\,\text{N}\), only this extra force changes the acceleration.
- Using Newton’s second law: \[ \Delta a = \frac{\Delta F}{m} = \frac{80}{10} = 8 \,\text{m s}^{-2} \]
- Hence, the increase in acceleration is \(8 \,\text{m s}^{-2}\).

- Apply work–energy theorem: work done equals change in kinetic energy.
- Work done by the variable force: \[ W = \int_{1.5}^{2.5} (-25x)\,dx = -50\,\text{J} \]
- Initial and final kinetic energies: \[ KE_1 = \frac{1}{2}m(6)^2 = 18m,\quad KE_2 = \frac{1}{2}m(4)^2 = 8m \]
- Change in kinetic energy: \[ \Delta KE = -10m \]
- Equating: \[ -50 = -10m \Rightarrow m = 5\,\text{kg} \]

- Initial kinetic energy: \[ KE = \frac{1}{2} \times 3 \times (10)^2 = 150\,\text{J} \]
- Potential energy at maximum height: \[ PE = 3 \times 10 \times 4.7 = 141\,\text{J} \]
- Energy lost: \[ \text{Loss} = 150 - 141 = 9\,\text{J} \]

- Diagonal of square \(= 80\,\text{cm}\) ⇒ side \(= 40\sqrt{2}\,\text{cm}\).
- Coordinates: \( A(0,0), B(40\sqrt{2},0), C(40\sqrt{2},40\sqrt{2}), D(0,40\sqrt{2}) \).
- Center of mass: \[ x_{\text{cm}} = 15\sqrt{2}, \quad y_{\text{cm}} = 15\sqrt{2} \]
- Distance from corner \(A\): \[ \sqrt{(15\sqrt{2})^2 + (15\sqrt{2})^2} = 30\,\text{cm} \]