A Spanish galleon is about to be boarded by bloodthirsty pirates in the shallows of a Caribbean island. To save a box of treasure on board, the captain orders his crew to secretly toss the box overboard, planning to come back for it later. The rectangular box is waterproof and measures 40.0 cm by 25.0 cm by 30.0 cm. It is made of wood and has mostly gold pieces inside, resulting in an average box density three times that of seawater.
Sinking below the surface, the box moves at constant vertical velocity of 1.15 m/s for 12.0 m before hitting the bottom. (a) Draw the free-body diagram for the box, (b) determine the magnitudes of the forces on the box, and (c) calculate the work done by each force and the net work done on the box. (d) Calculate the change in the box’s gravitational potential energy. (e) What is the change in the box’s total energy ?
(a) The free-body diagram for the box can be drawn as follows:
- A downward arrow representing the force of gravity acting on the box (mg), where m is the mass of the box and g is the acceleration due to gravity.
- An upward arrow representing the buoyant force (Fb), which is equal to the weight of the displaced fluid.
(b) The magnitudes of the forces on the box are:
- The force of gravity (mg)
- The buoyant force (Fb)
(c) The work done by each force and the net work done on the box can be calculated as follows:
- The work done by the force of gravity is given by the equation Wg = mgd, where d is the displacement of the box (12.0 m in this case).
- The work done by the buoyant force is zero since the displacement of the box is vertical and the buoyant force is perpendicular to the displacement.
- The net work done on the box is equal to the work done by the force of gravity.
(d) The change in the box's gravitational potential energy can be calculated using the equation ΔPE = mgh, where h is the change in height of the box. Since the box sinks to the bottom, h is equal to the depth of the water (12.0 m).
(e) The change in the box's total energy is equal to the sum of the change in gravitational potential energy and the work done on the box.
To solve this problem, let's break it down step by step:
(a) Draw the free-body diagram for the box:
A free-body diagram is a visual representation of the forces acting on an object. In this case, the box is sinking, so the only forces we need to consider are gravity and buoyancy. The weight of the box, which is equal to the force of gravity acting on it, points downward. The buoyant force, which opposes the weight, points upward. So, on the free-body diagram, you would draw an arrow pointing downward for the weight and an arrow pointing upward for the buoyant force.
(b) Determine the magnitudes of the forces on the box:
The weight of an object can be calculated using the formula: weight = mass * gravity. The mass can be calculated using the formula: mass = density * volume. In this case, the density of the box is three times that of seawater, so it is 3 times 1000 kg/m^3, which is equal to 3000 kg/m^3. The volume of the box can be calculated using the formula: volume = length * width * height. Plugging in the given values for length, width, and height, we get: volume = 0.4 m * 0.25 m * 0.3 m = 0.03 m^3.
So, the mass of the box is: mass = density * volume = 3000 kg/m^3 * 0.03 m^3 = 90 kg.
To calculate the weight of the box, we need to know the acceleration due to gravity. Let's assume it is 9.8 m/s^2. So, the weight of the box is: weight = mass * gravity = 90 kg * 9.8 m/s^2 = 882 N.
The buoyant force can be calculated using the formula: buoyant force = density_water * volume * gravity. The density of seawater is approximately 1000 kg/m^3. So, the buoyant force is: buoyant force = 1000 kg/m^3 * 0.03 m^3 * 9.8 m/s^2 = 294 N.
(c) Calculate the work done by each force and the net work done on the box:
To calculate the work done by a force, you can use the formula: work = force * distance * cos(theta), where theta is the angle between the force vector and the displacement vector. In this case, the box is moving vertically downward, so the angle between the weight vector and displacement vector is 180 degrees, and the angle between the buoyant force vector and the displacement vector is 0 degrees (since they are in the same direction).
The distance the box travels is given as 12.0 m. So, the work done by the weight is: work_weight = weight * distance * cos(180 degrees) = 882 N * 12.0 m * cos(180 degrees) = -10584 J (negative sign indicates that the work is done opposite to the direction of displacement).
The work done by the buoyant force is: work_buoyant = buoyant force * distance * cos(0 degrees) = 294 N * 12.0 m * cos(0 degrees) = 3528 J.
The net work done on the box is simply the sum of the work done by each force: net work = work_weight + work_buoyant = -10584 J + 3528 J = -7056 J.
(d) Calculate the change in the box's gravitational potential energy:
The change in gravitational potential energy can be calculated using the formula: change in PE = weight * change in height. In this case, the box moves downward by 12.0 m. So, the change in PE is: change in PE = weight * change in height = 882 N * (-12.0 m) = -10584 J.
(e) Calculate the change in the box's total energy:
Total energy is the sum of potential energy (PE) and kinetic energy (KE). Since the box moves at a constant velocity (1.15 m/s), its kinetic energy does not change. Therefore, the change in total energy is equal to the change in potential energy.
So, the change in the box's total energy is: change in total energy = change in PE = -10584 J.
In summary:
(a) Draw the free-body diagram for the box.
(b) Determine the magnitudes of the forces on the box: weight is 882 N, buoyant force is 294 N.
(c) Calculate the work done by each force: work done by weight is -10584 J, work done by buoyant force is 3528 J. The net work done on the box is -7056 J.
(d) Calculate the change in the box's gravitational potential energy: -10584 J.
(e) Calculate the change in the box's total energy: -10584 J.