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A ball of mass 10 kg and density 1 g/cm³ is attached to the base of a container by a spring (spring constant k = 200 N/m). The container is filled with a liquid of density 1.1 g/cm³. The entire system accelerates upward at 2 m/s². Find the elongation (extension) of the spring. (Take g = 10 m/s²)
- 2 cm
- 4 cm
- 6 cm
- 8 cm
Correct answer: 6 cm
Solution
Volume of ball V = m/rho_ball = 10 kg / 1000 kg/m³ = 0.01 m³. Effective g = 12 m/s². Buoyancy B = 1100 * 0.01 * 12 = 132 N. Weight W = 10 * 12 = 120 N. Spring force T = B - W = 12 N (spring stretched). Elongation x = T/k = 12/200 = 0.06 m = 6 cm.
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