Ch 07: Newton's Third Law

Knight Calc - Physics for Scientists and Engineers 5th Edition

Knight Calc5th EditionPhysics for Scientists and EngineersISBN: 9780137344796Not the one you use?Change textbook

Knight Calc 5th Edition

Ch 07: Newton's Third Law

Problem 25

Chapter 7, Problem 25

Two blocks are attached to opposite ends of a massless rope that goes over a massless, frictionless, stationary pulley. One of the blocks, with a mass of 6.0 kg, accelerates downward at (3/4)g. What is the mass of the other block?

Verified step by step guidance

Step 1: Begin by identifying the forces acting on each block. For the 6.0 kg block, the forces are its weight (gravitational force) and the tension in the rope. The downward acceleration is given as (3/4)g, where g is the acceleration due to gravity.

Step 2: Write the equation of motion for the 6.0 kg block using Newton's second law: $ F = ma $. The net force acting on the block is $ F_{net} = m_1 g - T $, where $ m_1 $ is the mass of the block, $ g $ is the acceleration due to gravity, and $ T $ is the tension in the rope. Substitute the acceleration $ a = \frac{3}{4}g $ into the equation: $ m_1 g - T = m_1 \cdot \frac{3}{4}g $.

Step 3: Solve for the tension $ T $ in the rope from the equation derived in Step 2. Rearrange the equation: $ T = m_1 g - m_1 \cdot \frac{3}{4}g $. Simplify to find $ T = \frac{1}{4}m_1 g $.

Step 4: Write the equation of motion for the second block (unknown mass $ m_2 $) using Newton's second law. Since this block is accelerating upward at $ \frac{3}{4}g $, the net force is $ T - m_2 g = m_2 \cdot \frac{3}{4}g $. Substitute $ T = \frac{1}{4}m_1 g $ into this equation: $ \frac{1}{4}m_1 g - m_2 g = m_2 \cdot \frac{3}{4}g $.

Step 5: Solve for $ m_2 $ by rearranging the equation from Step 4. Combine like terms and isolate $ m_2 $: $ \frac{1}{4}m_1 g = m_2 g + m_2 \cdot \frac{3}{4}g $. Factor out $ m_2 $ and simplify: $ m_2 = \frac{1}{4}m_1 $. Substitute $ m_1 = 6.0 \, \text{kg} $ to find the mass of the second block.

Verified video answer for a similar problem:

This video solution was recommended by our tutors as helpful for the problem above.

Video duration:

12m

Was this helpful?

Key Concepts

Here are the essential concepts you must grasp in order to answer the question correctly.

Newton's Second Law of Motion

Newton's Second Law states that the acceleration of an object is directly proportional to the net force acting on it and inversely proportional to its mass. This relationship is expressed by the equation F = ma, where F is the net force, m is the mass, and a is the acceleration. In this scenario, the forces acting on the blocks must be analyzed to determine the mass of the second block.

Tension in a Rope

In a system involving a pulley and a rope, tension is the force transmitted through the rope when it is pulled tight by forces acting at each end. Since the pulley is massless and frictionless, the tension is the same on both sides of the rope. Understanding how tension relates to the weights of the blocks and their acceleration is crucial for solving the problem.

Acceleration due to Gravity

Acceleration due to gravity (g) is the acceleration experienced by an object when it is in free fall near the Earth's surface, approximately 9.81 m/s². In this problem, the block's acceleration is given as 3/4g, which indicates that it is accelerating downward at a rate less than free fall. This value is essential for calculating the net force acting on the blocks and ultimately finding the mass of the second block.

Recommended video:

Guided course

05:20

Acceleration Due to Gravity

Related Practice

Textbook Question

A mobile at the art museum has a 2.0 kg steel cat and a 4.0 kg steel dog suspended from a lightweight cable, as shown in FIGURE EX7.21. It is found that $theta sub 1$ = 20° when the center rope is adjusted to be perfectly horizontal. What are the tension and the angle of rope 3?

views

Textbook Question

An 85 kg cheerleader stands on a scale that reads in kg. What does the scale read if the 85 kg cheerleader lifts the 50 kg cheerleader upward with an acceleration of 2.0 m/s²?

views

Textbook Question

The 100 kg block in FIGURE EX7.24 takes 6.0 s to reach the floor after being released from rest. What is the mass of the block on the left? The pulley is massless and frictionless.

views

Textbook Question

Your forehead can withstand a force of about 6.0 kN before fracturing, while your cheekbone can withstand only about 1.3 kN. Suppose a 140 g baseball traveling at 30 m/s strikes your head and stops in 1.5 ms. What is the magnitude of the force that stops the baseball?

views

Textbook Question

The 1.0 kg block in FIGURE EX7.23 is tied to the wall with a rope. It sits on top of the 2.0 kg block. The lower block is pulled to the right with a tension force of 20 N. The coefficient of kinetic friction at both the lower and upper surfaces of the 2.0 kg block is μ_k = 0.40. What is the tension in the rope attached to the wall?

views

Textbook Question

A rope of length L and mass m is suspended from the ceiling. Find an expression for the tension in the rope at position y, measured upward from the free end of the rope.

views