4.71 A block of mass \(20.0 \mathrm{~kg}\) supported by a vertical massless cable is initially at rest. The block is then pulled upward with a constant acceleration of \(2.32 \mathrm{~m} / \mathrm{s}^{2}\). a) What is the tension in the cable? b) What is the net force acting on the mass? c) What is the speed of the block after it has traveled \(2.00 \mathrm{m?}\)

Understanding the concept of tension in a cable is essential when dealing with mechanical systems that involve pulling or supporting masses. Tension can be thought of as the force conducted along the length of a cable or rope, acting in a direction that pulls on objects the cable is attached to. The tension in the cable must be sufficient to both support the weight of the mass and provide any additional force to accelerate the mass, if necessary.

When a block is suspended and accelerated upwards, the tension in the cable not only has to counteract the force of gravity but to also accelerate the block upwards. This can be mathematically expressed as:
\( T = m \times g + m \times a\), where \(T\) is the tension, \(m\) the mass of the block, \(g\) the acceleration due to gravity, and \(a\) the acceleration of the block. Here, it's critical to ensure the mass is properly accounted for when both gravity and acceleration are involved in the net force calculation.

The net force calculation is a fundamental aspect of Newton's second law, which states that the net force acting on an object is equal to its mass times its acceleration (\(F_{net} = m \times a\)). To visualize this, imagine the conflicting forces acting on an object: in one direction, you have weight due to gravity, and in another, you might have tension or some other applied force.

The net force is effectively the vector sum of all these forces acting on an object. If the mass is stationary, the net force is zero, implying a state of equilibrium. However, if the object is accelerating, the net force will be non-zero and in the direction of the acceleration. In the context of the given exercise, the upward tension and downward gravitational force were considered to find the net force acting on the block.

The kinematic equations are a set of four equations that describe the motion of objects under constant acceleration. These equations connect the five kinematic variables: displacement, initial velocity, final velocity, acceleration, and time.

For situations like our exercise, where an object starts from rest and is subject to a consistent acceleration, one of the key equations we use is \( v^2 = u^2 + 2 \times a \times d \), where \( v \) is the final velocity, \( u \) the initial velocity (zero in cases starting from rest), \( a \) the acceleration, and \( d \) the displacement. Understanding these kinematic equations is crucial for problems involving motion, as they provide a reliable method to calculate various motion parameters from known values.

An object experiences constant acceleration when it speeds up or slows down at a steady rate over time. In physics, this implies that the object's acceleration stays the same in magnitude and direction.

When solving physics problems, constant acceleration can simplify calculations because it allows us to use the kinematic equations confidently. For instance, in the provided exercise, knowing that the block is pulled upward with a constant acceleration ensures that we can predict its movement accurately over any given distance or time interval. This property is particularly useful as it makes the system's behavior more predictable and thus easier to study and understand.

The final velocity of an object is the speed and direction it has after undergoing acceleration over a distance or period. It can be computed using kinematic equations, and in scenarios with constant acceleration, the calculation is straightforward.

In our case, the block's final velocity after traveling a certain distance was determined using the equation \( v^2 = u^2 + 2 \times a \times d \). Key takeaways include that if an object starts from rest, the initial velocity (\(u\)) is zero, making the equation simpler. Moreover, the square root must be taken after plugging in the values to solve for \(v\). Understanding how to calculate final velocity is critical in predicting future positions and speeds of objects in motion.