A box with a volume V = 0.04 m3 lies at the bottom of a lake whose water has a density of 103 kg/m3.

(a) How much force is required to lift the box, if the mass of the box is 1050 kg?


(b) How much force is required to lift the box, if the mass of the box is 100 kg?


(c) How much force is required to lift the box, if the mass of the box is 70 kg?

figure it out

force=mg(mass*acceleration due to gravity

To find the force required to lift the box, we need to consider the buoyant force acting on it when submerged in water.

The buoyant force is equal to the weight of the water displaced by the object. It can be calculated using Archimedes' principle, which states that the buoyant force is equal to the weight of the fluid displaced by the submerged object.

The volume of the box is given as V = 0.04 m^3. Since the box is fully submerged in the water, the volume of water displaced is equal to the volume of the box itself.

(a) For the box with a mass of 1050 kg:
The weight of the water displaced by the box is equal to the density of water (103 kg/m^3) times the volume of the box.
Weight of water displaced = density * volume = 103 kg/m^3 * 0.04 m^3 = 4.12 kg

The buoyant force acting on the box is equal to the weight of the water displaced:
Buoyant force = weight of water displaced = 4.12 kg * 9.8 m/s^2 = 40.376 N

To lift the box, a force equal to the buoyant force needs to be applied in the opposite direction:
Force required = 40.376 N

(b) For the box with a mass of 100 kg:
Following the same calculation, the buoyant force will be:
Buoyant force = 103 kg/m^3 * 0.04 m^3 * 9.8 m/s^2 = 40.376 N

Hence, the force required to lift the box is also 40.376 N.

(c) For the box with a mass of 70 kg:
Again, using the same calculation, the buoyant force will be:
Buoyant force = 103 kg/m^3 * 0.04 m^3 * 9.8 m/s^2 = 40.376 N

Therefore, the force required to lift the box in this case is also 40.376 N.