Problem 17 - University Physics with Modern Physics

Question

A light rope is attached to a block with mass 4.00 kg that rests on a frictionless, horizontal surface. The horizontal rope passes over a frictionless, massless pulley, and a block with mass \(m\) is suspended from the other end. When the blocks are released, the tension in the rope is 10.0 \(\mathrm{N}\) (a) Draw two free-body diagrams, one for the \(4.00-\) kg block and one for the block with mass \(m\) . (b) What is the acceleration of either block? (c) Find the mass \(m\) of the hanging block. (d) How does the tension compare to the weight of the hanging block?

Step-by-Step Solution

Answer

The acceleration is 2.5 m/s², mass of the hanging block is 1.37 kg, and tension is less than its weight.

Step 1: Analyze the Problem

We are dealing with a system of two blocks connected by a rope over a pulley. One block is on a frictionless horizontal surface, and the other is hanging vertically. We need to find the tension in the rope and the acceleration of the blocks.

Step 2: Draw Free-Body Diagram for 4.00 kg Block

For the block on the horizontal surface, draw a free-body diagram. This block has a mass of 4.00 kg, and the only horizontal force acting on it is the tension in the rope, which is 10.0 N. There are no vertical forces acting because it is on a horizontal surface.

Step 3: Draw Free-Body Diagram for Block with Mass m

For the hanging block, draw a free-body diagram. This block is subjected to two forces vertically: its weight, which is the gravitational force \( W = mg \), and the tension in the rope which is acting upwards with a force of 10.0 N.

Step 4: Write Equations of Motion for 4.00 kg Block

Since the surface is frictionless, the equation of motion for the horizontal block is given by Newton's Second Law: \( T = 4.00 \, \text{kg} \times a \), where \( T = 10.0 \, \text{N} \). So, \( a = \frac{10.0 \, \text{N}}{4.00 \, \text{kg}} \).

Step 5: Calculate the Acceleration

Calculate the acceleration of the 4.00 kg block using the formula \( a = \frac{T}{m} = \frac{10.0 \, \text{N}}{4.00 \, \text{kg}} = 2.5 \, \text{m/s}^2 \). This is the same acceleration for the system since they are connected.

Step 6: Write Equation of Motion for Hanging Block

For the hanging block, apply Newton's Second Law: \( mg - T = ma \). Given \( T = 10.0 \, \text{N} \), substitute \( a = 2.5 \, \text{m/s}^2 \) and solve for \( m \): \( mg = 10.0 \, \text{N} + ma \).

Step 7: Solve for Mass m

Use the equation from Step 6: \( m \cdot 9.8 \, \text{m/s}^2 = 10.0 \, \text{N} + m \cdot 2.5 \, \text{m/s}^2 \). Simplify \( m \cdot (9.8 - 2.5) = 10.0 \). Solve to get \( m = \frac{10.0}{7.3} \approx 1.37 \, \text{kg} \).

Step 8: Compare Tension and Weight of Hanging Block

The weight of the hanging block is \( mg \approx 1.37 \, \text{kg} \times 9.8 \, \text{m/s}^2 \approx 13.43 \, \text{N} \). The tension in the rope is 10.0 N, which is less than the weight of the hanging block (13.43 N), indicating that the tension is not enough to fully support the weight, leading to the acceleration.

Key Concepts

tension in the rope

In this exercise, we explore the concept of tension through a system involving two blocks and a rope. When it comes to understanding tension in the rope, it helps to visualize what's happening. Imagine the rope as linking two blocks: one lying on a frictionless table and the other hanging from a pulley. Simply put, tension is the force transmitted through the rope from one block to the other.

In an ideal scenario without friction or external forces interfering, the tension can be considered equal throughout the rope. That's because the rope is massless, so it distributes the tension evenly from one end to the other. Here, the tension has been measured as 10.0 N. This means that a force of 10.0 N is consistently being applied at both ends of the rope - on the block resting on the table and the one hanging.

Understanding tension helps in solving many physics problems involving ropes, cables, or chains. It's important to remember that tension is not a push, but a pull. It tries to "stretch" the rope apart. We'll also see how tension relates with other forces affecting the blocks, such as gravity, in applying Newton’s Second Law.

acceleration of blocks

Acceleration is a key factor in understanding how forces influence the motion of objects. In our problem, both blocks are accelerating at the same rate because they are connected by the rope. Let's break down how we determine their shared acceleration.

For the block on the frictionless table, the only horizontal force applied is the tension in the rope (10.0 N). According to Newton’s Second Law, force equals mass times acceleration: \[ F = ma \]Here, we can rearrange that to find acceleration:\[ a = \frac{F}{m} \]Where the force is 10.0 N and the mass of the block is 4.00 kg. This gives:\[ a = \frac{10.0 \, \text{N}}{4.00 \, \text{kg}} = 2.5 \, \text{m/s}^2 \]

The acceleration of 2.5 m/s² is the rate at which both blocks change their velocity because the tension pulls them towards each other. This acceleration acts downward for the hanging block and horizontally for the block on the table. Whenever you have connected systems like this, their accelerations will be shared and are fundamental in analyzing the dynamic behavior of each component involved.

Newton's Second Law

Newton's Second Law is like the backbone of motion problems, particularly those involving forces we can't directly control. This law states that the acceleration of an object depends directly on the net force acting on it and inversely on its mass. In simpler terms, the more force you apply, the faster an object will accelerate. Conversely, the heavier the object, the more force you'll need to change its motion.Let's apply this to both of our blocks:- For the 4.00 kg block on the horizontal surface: It's primarily affected by the rope's tension. The net force here is just the tension (10.0 N), leading to the equation: \[ T = m a = 4.00 \, \text{kg} \times a \] Solving gives us the acceleration: \( a = 2.5 \, \text{m/s}^2 \).- For the hanging block with mass \( m \): It experiences both the force of gravity \( mg \) and the upward tension of 10.0 N. The net force here can be described by: \[ mg - T = ma \] Solving this equation helps us find that \( m \approx 1.37 \, \text{kg} \).

Newton's Second Law simplifies to the real-world pressing question of: "How much force does it take to move this object?" It’s crucial in understanding why these blocks move the way they do, helping us predict not only the acceleration but, through rearranging the equations, solve for unknowns like mass or force.