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A free-body diagram is a crucial tool in physics that helps us visualize the forces acting on an object. In the case of a spaceship coming in for a vertical landing, we focus on two main forces:

• Gravitational Force (Weight): This force pulls the spaceship downward towards the Earth. It is due to gravity and is calculated as the product of the mass and gravitational acceleration, given in the problem as 75,600 N.
• Braking Force: Acting upwards, this force is exerted by the rockets to counteract the pull of gravity and slow the spaceship down.

By illustrating these forces accurately on the diagram, with arrows representing their magnitude and direction, we gain better insights into the problem solving process. In our diagram, the gravitational force would be shown as an arrow pointing downward (Earth's gravity), and the braking force would be an arrow pointing upward against gravity.

Uniform acceleration refers to the scenario where an object's velocity changes at a constant rate. In the exercise, the spaceship decelerates uniformly, meaning it comes to a stop following a steady decrease in velocity over time. This concept is essential when studying motion, as it allows us to use specific kinematic equations to solve for unknown variables like acceleration or final velocity.
In our context, the spaceship starts at an initial speed of 1.00 km/s, and comes to rest (0 m/s) in 120 seconds (2 minutes). To find the acceleration, we use the equation:
\[ v_f = v_i + at \]
Here, \( a \) is the uniform acceleration, \( v_f \) is the final velocity (0 m/s), \( v_i \) is the initial velocity (1000 m/s), and \( t \) is the time (120 s). Solving this gives \( a = -8.333 \, \text{m/s}^2 \). This negative sign indicates that the acceleration is in the opposite direction of initial velocity, i.e., it is a deceleration.

Newton's Second Law of Motion is a fundamental principle that describes how the motion of an object is affected by the forces acting upon it. It can be mathematically expressed as:
\[ F_{net} = ma \]
where \( F_{net} \) is the net force acting on an object, \( m \) is its mass, and \( a \) is the acceleration. In this problem, the net force is calculated as the difference between the upward braking force \( F_b \) provided by the spaceship's rockets, and the downward gravitational force \( F_g \).
When considering the spaceship:

• Mass Calculation: The mass \( m \) can be derived from the gravitational force: \( F_g = mg \). By rearranging, \( m = \frac{F_g}{g} \), where \( g \) is the acceleration due to gravity (9.81 m/s²).
• Determining Braking Force: With the mass and acceleration known, we substitute into the equation: \( F_b = ma + F_g \), and solve for \( F_b \) to find the requisite braking force.

Thus, this law provides a logical pathway for solving for forces in the system by linking force, mass, and acceleration.

Calculating the braking force is central to ensuring that the spaceship comes to a complete and safe stop during its vertical landing. Using the formulas derived from the exercise, especially Newton's Second Law, we can compute the exact force needed from the spaceship's rockets.
First, we find the mass of the spaceship using the gravitational force:\
\[ m = \frac{F_g}{g} = \frac{75600}{9.81} \approx 7708 \, \text{kg} \]
Then, we plug the mass and the previously calculated deceleration into the net force equation. We express the braking force \( F_b \) as:
\[ F_b = ma + F_g \]
Substitute the values:
\[ F_b = 7708 \times (-8.333) + 75600 \]
\[ F_b = -64233.764 + 75600 \]
\[ F_b \approx 11366 \, N \]
This force is in the upward direction, counterbalancing gravity to bring the spaceship to rest. Calculating it accurately ensures the spaceship's safe landing by applying the right amount of thrust against gravity.