A 5.00 kg object is released from rest while fully submerged in a liquid. The liquiddisplaced by the submerged object has a mass of 3.00 kg. How far and in what directiondoes the object move in 0.200 s, assuming that it moves freely and that the drag force onit from the liquid is negligible?

To solve this problem, we need to understand the forces acting on the object. The two main forces are the buoyant force (upward) and the gravitational force (downward).

1. Calculate the gravitational force (weight of the object): The gravitational force can be calculated using the formula F = m*g, where m is the mass of the object and g is the acceleration due to gravity. In this case, m = 5.00 kg and g = 9.81 m/s² (approximate value on Earth's surface). So, F_gravity = 5.00 kg * 9.81 m/s² = 49.05 N.

2. Calculate the buoyant force (weight of the displaced liquid): The buoyant force is equal to the weight of the liquid displaced by the object. This can be calculated using the same formula as above, but with the mass of the liquid instead of the object. So, F_buoyancy = 3.00 kg * 9.81 m/s² = 29.43 N.

3. Determine the net force on the object: The net force is the difference between the buoyant force and the gravitational force. Since the buoyant force is less than the gravitational force, the net force is downward. F_net = F_gravity - F_buoyancy = 49.05 N - 29.43 N = 19.62 N.

4. Use Newton's second law to find the acceleration of the object: Newton's second law states that F = m*a, where F is the net force, m is the mass of the object, and a is the acceleration. Solving for a gives a = F/m = 19.62 N / 5.00 kg = 3.924 m/s².

5. Use the kinematic equation to find the distance the object moves in 0.200 s: The equation d = 0.5at² can be used, where d is the distance, a is the acceleration, and t is the time. So, d = 0.5 * 3.924 m/s² * (0.200 s)² = 0.157 m.

So, the object moves 0.157 m downward in 0.200 s.