Question

Chapter 14, Problem 030 A 5.75 kg object is released from rest while fully submerged in a liquid. The liquid displaced by the submerged object has a mass of 2.62 kg. Take the upward direction to be positive. What is the displacement of the object in 0.151 s, assuming that it moves freely and that the drag force on it from the liquid is negligible? Number Units the tolerance is +/-2% Click if you would like to Show Work for this question: Open Show Work

Answer 1

here,

Since the object is fully submerged, the volume of the object is the volume of the displaced liquid. The buoyant force is equal the weight of the displaced liquid.

Buoyant force = 2.62 * 9.8 = 35.676 N

Weight of the object = 5.75 * 9.8 = 56.35 N

Net force = 56.35 – 35.676 = 20.67 N

a = net force /mass = 3.6 m/s^2

Since the weight is greater than the buoyant force, the object is moving downward.

the displacement of object , s = ( u * t - 0.5 * a * t^2)

s = ( 0 - 0.5 * 3.6 * 0.151^2) m

s = - 0.04 m