First, we need to determine what the apparent weight actually is. If we draw a free body diagram of the weights, we note that the force due to gravity acts downward and the force of the platform on the weights (the normal force) acts upward.

The normal force, F_N in the above diagram, is what a scale reads, and is thus the apparent weight. Finding the normal force is not as easy as equating it to the weight because the system is in free-fall, and thus accelerating. We can use Newton’s second law and equate it to acceleration to find the normal force.

Assuming downward is the positive y direction, we can solve for F_N.

mg – F_N = ma

F_N = mg – ma = 0 N

Logically, this makes sense. If an object is in free fall, the acceleration of the system is the same as the acceleration due to gravity. 

The acceleration of gravity is given by the equation , where G is constant.

For Earth, .

For the new planet, 

.

So, the acceleration is the same in both cases.

The force between two objects is calculated by:

Use our given variables:

The best way to think of a system in equilibrium is that it is displaying a constant velocity in both the horizontal and vertical directions. Both the skydiver and the tightrope walker have a constant velocity. The driver, however, is driving around a circular track. You must remember that velocity is a vector of magnitude and direction. Because the direction of the driver is changing, he is not driving at a constant velocity; thus, he is not in equilibrium, and is experiencing an acceleration (centripetal acceleration).

Keep in mind that the skydiver is experiencing both the downward force of gravity, and the upward force of air resistence on his deployed parachute. These forces cancel, producing a net force of zero, and allowing him to fall at a constant velocity in equilibrium.

Since the only acceleration on the object is to the west, we can conclude that there is no vertical acceleration on the object. As a result, the northern force must be offset by the third force. The only way to offset the northern force is to have a force pointing south, in any direction. As such, the third force could be directly south, southwest, or southeast as long as the southern component can counteract the northern force.

The third force would only need to be equal to the northern force if it was pointing directly due south, and we can only confirm that the third force is not completely offsetting the western force. As a result, we can only conclude that the third force has some magnitude in the southern direction.

The skydiver’s velocity (along the y-axis) will be zero when he initially jumps out of the plane, however, his net force will be greatest at that point.

As the skydiver first jumps, there is only the force of gravity acting on him. As he approaches terminal velocity, however, the force of wind resistance becomes equal and opposite to the force of gravity, and he begins to move at constant velocity with a net force of zero.

The velocity of the mass will be the greatest at the equilibrium position. Although the restoring force on the mass due to the spring will be the greatest at maximum displacement from the equilibrium position, this force is being applied to the mass as it moves in the opposite direction. As the mass moves towards the equilibrium position from maximum displacement, the restoring force due to the spring is applied in the same direction as the velocity of the mass. Because of this, the object is accelerating. After the mass passes through the equilibrium position, the spring begins to apply a force in the direction opposite to the motion of the mass, resulting in the deceleration of the mass. The mass experiences its maximum velocity at the equilibrium position.

When an object is decelerating, there is a net force in the opposite direction of the motion of the object. Usually, this force is friction, which always acts counter to the net velocity. Even if there is a force in the direction of the object's motion, the net force is backwards if the object is decelerating.

Mathematically, we can prove this concept using Newton's second law. The direction of the object's motion will be positive, while the opposite (backwards) direction will be negative.

If the object is decelerating, then the acceleration is negative.

From this calculation, the force must be negative, acting counter to the direction of motion.

When an object is at terminal velocity, the upward force on the object from air resistance is equal to the force of gravity. These equal, but opposing, forces cancel each other out and the net force is zero. This results in a zero acceleration, allowing the object to fall at a constant (terminal) velocity.

Consider that tension is a contact force that acts over a distance. Assuming the rope does not stretch and has no mass (as we should for the MCAT), we can think of tension as a force that acts directly on the mass. We can draw a force diagram below.

If the system is stationary (at equilibrium), we can see that tension is equal to the weight, T = mg.

T = (2 kg)(9.8 m/s²) = 19.6N