A wooden block of mass 0.5 kg and density 800 kg m ⁻³ is attached to the free end of a vertical spring having a spring constant of 50 N m ⁻¹ , which is fixed at the bottom. The entire system is completely submerged in water. Determine the elongation (or compression) of the spring in equilibrium.

Step-by-step solution

Volume of the wooden block is V = m _ b = 0.5 800 m ³ Buoyant force acting on the block in water is F_ B = V _ w g Substituting the values, F_ B = ( 0.5 800 ) 1000 10 = 6.25 N Weight of the block is W = mg = 0.5 10 = 5 N Since F_ B > W , the net force on the block is directed upwards. To maintain equilibrium, the spring must exert a downward force, which means it will be elongated. Let x be the elongation of the spring. At equilibrium, F_ B = W + kx 6.25 = 5 + 50x 50x = 1.25 x = 1.25 50 m = 0.025 m x = 2.5 cm