Problem 49 - College Physics

Question

A \(1.00-\) kg block \((M)\) is on a flat frictionless surface (vFig. 5.32). This block is attached to a spring initially at its relaxed length (spring constant is \(50.0 \mathrm{~N} / \mathrm{m}\) ). A light string is attached to the block and runs over a frictionless pulley to a \(450-\mathrm{g}\) dangling mass \((m)\). If the dangling mass is released from rest, how far does it fall before stopping?

Step-by-Step Solution

Answer

The mass falls approximately 6.08 cm before stopping.

Step 1: Identifying Forces

First, identify the forces acting on each mass. The dangling mass, \(m\), experiences gravitational force \(F_g = mg\), where \(g = 9.8 \, \text{m/s}^2\). The block \(M\) is connected to the spring and a pulley system; the force exerted by the spring is \(F_{spring} = kx\), where \(k = 50.0 \, \text{N/m}\) and \(x\) is the displacement.

Step 2: Setting up Equations

Apply Newton’s second law to each mass. For the block \(M\) on the surface, \(M a = kx\). For the dangling mass \(m\), the equation is \(m a = mg - T\), where \(T\) is the tension in the string. Set \(T = kx\) because tension is equal to the spring force when the block is in motion by displacement \(x\).

Step 3: Solving for Acceleration

Combine the equations: \(ma = mg - kx\) and \(Ma = kx\). Substitute the values to get \(0.45 a = 0.45 \times 9.8 - 50x\) and \(a = \frac{kx}{M}\). Solve for \(a\) using these equations where both acceleration \(a\) and displacement \(x\) are involved.

Step 4: Solving for Displacement

Solve the system of equations for \(x\) and \(a\). The simplified equation to find the displacement is \((m + M) a = mg\), substituting \(a = \frac{kx}{M}\). Solving gives \(x = \frac{mgM}{k(m+M)}\). Substitute values: \(x = \frac{0.45 \times 9.8 \times 1.00}{50 \, (0.45 + 1.00)}\). This gives the displacement necessary to stop the mass.

Step 5: Calculating Displacement

Now compute the value of \(x\) using the simplified equation: \(x = \frac{4.41}{72.5} = 0.0608 \, \text{meters}\) or approximately \(6.08 \, \text{cm}\).

Key Concepts

Force and Motion

Understanding the relationship between force and motion is key to solving problems involving physical objects. In this exercise, Newton's second law, which states that the acceleration of an object is directly proportional to the net force acting on it, plays a pivotal role. The formula is expressed as: \( F = ma \), where \( F \) represents force, \( m \) is the mass, and \( a \) is the acceleration.

In the given problem, there are two masses: a block on a surface and a dangling mass connected by a pulley system. Each mass experiences different forces. The dangling mass is under the influence of gravity, which pulls it downward, while the block on the surface experiences force due to the spring. These forces affect acceleration and motion, demonstrating Newton's second law.

Spring Constant

The spring constant, denoted as \( k \), is a measure of a spring's stiffness. The larger the value of \( k \), the stiffer the spring. In our problem, we use Hooke's Law to determine the spring force: \( F_{spring} = kx \), where \( x \) is the displacement from the spring's natural length.

This problem involves a spring with a constant of \( 50.0 \, \text{N/m} \). As the dangling mass pulls on the string, it causes the spring to stretch. The spring force opposes the gravitational force pulling the mass down. Thus, understanding the spring constant helps us predict how much the spring will stretch, which is essential for calculating the final position of the masses.

Pulley System

A pulley system can change the direction of a force and make lifting easier. In this scenario, the pulley connects the two masses and ensures that they move in response to the same forces. Because we assume the pulley is frictionless, it doesn't consume energy or reduce the force transmitted through the string.

The tension in the string in this scenario is crucial. It connects the motion of the two masses: the block on the table and the dangling mass. The tension directly affects how these masses accelerate and move, allowing us to set up equations that describe their motion accurately.

Displacement Calculation

Displacement refers to how far an object has moved from its starting position. It is a vector quantity, meaning it has both magnitude and direction. In this problem, it is important to find the displacement of the dangling mass when it comes to rest.

Using the equations set from Newton's second law, the spring force, and tension, we can solve for the displacement. The step-by-step solution shows that the displacement \( x \) is calculated by rearranging and combining these formulas. Specifically, we derived that the expression for displacement is: \[ x = \frac{mgM}{k(m + M)} \]

• where \( m \) is the mass of the dangling object,
• \( g \) is the acceleration due to gravity,
• \( M \) is the mass of the block,
• and \( k \) is the spring constant.

By substituting the given values, we determined that the mass falls about \(6.08\) cm before stopping.