Question

A spring of negligible mass is attached to the ceiling of an elevator. When the elevator is stopped at the first floor, a mass \(M\) is attached to the spring, stretching the spring a distance \(D\) until the mass is in equilibrium. As the elevator starts upward toward the second floor, the spring stretches an additional distance \(D / 4\). What is the magnitude of the acceleration of the elevator? Assume the force provided by the spring is linearly proportional to the distance stretched by the spring.

Short answer

Answer: The magnitude of the acceleration of the elevator is equal to \(g/4\), where \(g\) is the gravitational acceleration.

Step-by-step solution

Apply Newton's Second Law at the initial equilibrium position

To analyze the forces on the mass when the elevator is at rest and the mass is hanging in equilibrium from the spring, we can use Newton's Second Law: $$ \sum F = M a $$ In this case, the forces acting on the mass are gravitational force (downward) and the spring force (upward). At the equilibrium position, the net force is zero and the acceleration is also zero: $$ F_{spring} - Mg = 0 $$

Apply Hooke's Law to find spring constant

According to Hooke's law, the force exerted by the spring is proportional to the distance it is stretched (\(F_{spring} = kD\)), where \(k\) is the spring constant. Therefore, the spring constant can be calculated as: $$ k = \frac{F_{spring}}{D} = \frac{Mg}{D} $$

Find the force exerted by the spring when the elevator accelerates

According to the problem, the spring stretches an additional distance of \(D/4\) as the elevator accelerates. The total distance stretched by the spring at that moment is \(D + D/4 = \frac{5D}{4}\). Using Hooke's Law, we can find the force exerted by the spring at this position: $$ F'_{spring} = k \frac{5D}{4} = \frac{5Mg}{4} $$

Apply Newton's Second Law when the elevator accelerates

In this case, we need to consider the net force acting on the mass during acceleration: $$ \sum F = M a $$ As the elevator accelerates upward, the forces acting on the mass are the gravitational force (downward) and the spring force (upward): $$ F'_{spring} - Mg = M a $$

Calculate the acceleration

Now, we can substitute the values of \(F_{spring} '\) and solve for the acceleration: $$ \frac{5Mg}{4} - Mg = M a $$ We can simplify the equation: $$ \frac{Mg}{4} = M a $$ Finally, we can solve for the acceleration \(a\): $$ a = \frac{g}{4} $$ So, the magnitude of the acceleration of the elevator is equal to \(g/4\).

Key concepts

Newton's Second Law

When it comes to understanding how objects move and interact, Newton's Second Law provides a fundamental foundation. The law states that the acceleration of an object is directly proportional to the net force acting on it and inversely proportional to its mass. In equation form, it is expressed as \[ F = ma \] where \(F\) represents the net force applied to the object, \(m\) is the mass of the object, and \(a\) is the acceleration. This equation is particularly useful in analyzing physics elevator problems, as it allows us to determine the elevator's acceleration based on the forces present.

In the case of our elevator, when stopped at the first floor, the equilibrium indicates that the spring force equals the gravitational force. As the elevator begins to move, the spring stretches further, implying that the net force is no longer zero. Using Newton's Second Law, we can isolate the acceleration and find out how quickly the elevator is moving upward. The beauty of this law is in its simplicity and wide applicability, making it a powerful tool in physics.

Hooke's Law

Understanding the behavior of springs is crucial when analyzing physics elevator problems. Hooke's Law describes this behavior by stating that the force needed to extend or compress a spring by some distance is proportional to that distance. Mathematically, Hooke's Law is represented as \[ F_{spring} = kx \] where \(F_{spring}\) is the force exerted by the spring, \(k\) is the spring constant, a measure of the spring's stiffness, and \(x\) is the distance the spring has stretched or compressed from its original length.

In our elevator scenario, when the mass reaches equilibrium, the spring's force balances the weight of the mass. Any additional stretch due to the elevator's motion can be analyzed using Hooke's Law to find the new force exerted by the spring. This law is vital for solving problems that involve spring forces and is intimately linked with the concept of simple harmonic motion.

Spring Constant

The spring constant, denoted as \(k\), is a parameter that quantifies the stiffness of a spring. A larger spring constant means a stiffer spring, which requires more force to stretch or compress a given distance. Conversely, a smaller spring constant indicates a more flexible spring. We calculate the spring constant by rearranging Hooke's Law as \[ k = \frac{F_{spring}}{x} \] where \(F_{spring}\) is the force the spring exerts, and \(x\) is the spring's displacement from its rest position.

Understanding the spring constant is essential in our elevator problem to determine how much the spring will stretch when additional forces are applied, such as those resulting from the elevator accelerating. Knowing the spring constant allows us to calculate the forces involved in the system.

Equilibrium

Equilibrium occurs when all the forces acting on an object are balanced, leading to a state where there is no net force and, consequently, no change in motion. In the context of our physics elevator problem, when the mass attached to the spring is at rest and the elevator is not moving, the spring reaches an equilibrium position. This means that the upward spring force precisely counterbalances the downward gravitational force on the mass. At equilibrium, the net force is zero, \[ F_{net} = 0 \] and the object is either at rest or moves with a constant velocity. An understanding of equilibrium is important for solving many physics problems because it establishes a baseline condition from which we can analyze changes in the system, such as the elevator starting to move.

Gravitational Force

Gravitational force, also known as weight, is the force with which the Earth attracts a body toward its center. It is given by the equation \[ W = mg \] where \(m\) is the mass of the body, and \(g\) is the acceleration due to gravity. In the elevator problem, the gravitational force is acting downward on the mass attached to the spring, and it is this force that causes the initial stretching of the spring to achieve equilibrium.

When the elevator accelerates upwards, the gravitational force remains constant, but the spring force changes due to the additional stretch. The interplay between the gravitational force and the spring force determines the motion and acceleration of the mass in the system. Gravitational force is significant not only because it explains the weight of objects but also because it plays a pivotal role in the analysis of any problem dealing with objects near Earth's surface.