How Much Force Does Water Apply to a Sandbag Dropped from a Hot-Air Balloon?

[Brittykitty]

At an altitude of 315 m above a lake, a hot-air balloon pilot drops a 2.50 kg sandbag. If the bag comes to a stop 4.50 m below the surface of the lake, what force was applied by the water on it?

Mass:2.50kg
velocity: 2.50kg*9.81m/s=61.31?

I can't remember how to figure out what the time would be?

 

[rock.freak667]

It's easier to use a conservation of energy method. So try that instead

 

[kerimek]

I would approach this problem using the principles of Newton's laws of motion. The initial force applied to the sandbag would be the force of gravity, which is equal to the mass of the sandbag (2.50 kg) multiplied by the acceleration due to gravity (9.81 m/s^2), resulting in an initial force of 24.525 N.

When the sandbag is dropped, it will experience a downward force due to gravity and an upward force due to the buoyant force of the water. As the sandbag falls through the water, it will also experience a drag force due to the resistance of the water. This drag force will increase as the sandbag's velocity increases.

Once the sandbag reaches the bottom of the lake, the upward buoyant force will equal the downward force of gravity, causing the sandbag to come to a stop. At this point, the force applied by the water on the sandbag will be equal and opposite to the force of gravity, which is 24.525 N.

To determine the time it takes for the sandbag to reach the bottom of the lake, we can use the equation for displacement: d = v0t + 1/2at^2, where d is the displacement (4.50 m), v0 is the initial velocity (which is 0 m/s since the sandbag is dropped), a is the acceleration (9.81 m/s^2), and t is the time. Rearranging the equation, we get t = √(2d/a) = √(2*4.50m/9.81m/s^2) = 0.96 s.

Therefore, the force applied by the water on the sandbag is 24.525 N and it takes 0.96 seconds for the sandbag to reach the bottom of the lake.