/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 122 On earth a block has a weight of... [FREE SOLUTION] | 91影视

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Chapter 4: Problem 122

On earth a block has a weight of \(88 \mathrm{N}\). This block is sliding on rope in which the tension is \(T=24 \mathrm{N}\). The coefficient of kinetic friction between the block and the surface is \(\mu_{k}=0.20 .\) Concepts: (i) Which of Newton's laws of motion provides the best way to determine the acceleration of the block? (ii) Does the net force in the \(x\) direction equal the tension \(T\) ? Calculations: Determine the acceleration of the block.

Short Answer

Expert verified
The acceleration of the block is approximately 0.71 m/s虏.

Step by step solution

01

Understanding the Physics Concepts

In this problem, Newton's Second Law provides the best way to determine the acceleration. It states that the sum of the forces acting on an object is equal to the mass of the object multiplied by its acceleration, expressed as \( F = ma \).
02

Analyzing Forces in the X Direction

The net force in the \( x \) direction is not equal to the tension \( T \) alone. It is the difference between the tension and the force of kinetic friction. Therefore, \( F_{net} = T - f_k \).
03

Calculating the Force of Kinetic Friction

The force of kinetic friction \( f_k \) is calculated using the formula \( f_k = \mu_k \cdot N \). Since the block is on a flat surface, the normal force \( N \) is equal to the weight of the block. So, \( f_k = 0.20 \times 88 \).
04

Determining the Net Force

Calculate \( f_k = 0.20 \times 88 = 17.6 \) N. Thus, the net force in the \( x \) direction is \( F_{net} = 24 - 17.6 = 6.4 \) N.
05

Finding the Mass of the Block

The weight of the block is the force due to gravity, \( W = mg \). Therefore, the mass \( m = \frac{W}{g} = \frac{88}{9.8} \approx 8.98 \) kg.
06

Calculating the Acceleration

Now use \( F = ma \) : \( 6.4 = 8.98 \cdot a \). Solve for \( a \) to find \( a = \frac{6.4}{8.98} \approx 0.71 \) m/s虏.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Kinetic Friction
When you slide a block along a surface, it encounters resistance. This resistance is due to kinetic friction, which acts opposite to the direction of motion. The force of kinetic friction is calculated using the formula \( f_k = \mu_k \cdot N \). Here, \( \mu_k \) is the coefficient of kinetic friction, characterizing the roughness of the surfaces in contact. In our exercise, \( \mu_k = 0.20 \).
The normal force \( N \) is usually equal to the weight of the block when it is on a flat surface. For a block weighing 88 N, this gives \( f_k = 0.20 \times 88 = 17.6 \) N. This value tells us how much force opposes the sliding block.
Net Force
To calculate the net force acting on the block, you need to consider both the tension in the rope and the kinetic friction force. Contrary to what might initially be expected, the net force \( F_{net} \) is not just the tension \( T = 24 \) N pulling the block.
Instead, it considers the opposing force of kinetic friction. This means \( F_{net} = T - f_k \). In our case, we have \( F_{net} = 24 - 17.6 = 6.4 \) N. This net force is crucial in determining the motion of the block.
Acceleration Calculation
Knowing the net force allows us to calculate the block's acceleration using Newton's Second Law, which is expressed by \( F = ma \). Here, \( F \) is the net force, \( m \) is the mass of the block, and \( a \) is the acceleration we want to find.
To get the mass \( m \), we use the relation \( W = mg \), where \( W \) is the weight of the block. This gives \( m = \frac{88}{9.8} \approx 8.98 \) kg. So, using the net force, the equation becomes \( 6.4 = 8.98 \cdot a \). Solving for \( a \), you find \( a = \frac{6.4}{8.98} \approx 0.71 \) m/s虏. This tells us how quickly the block is speeding up as it slides.
Block Dynamics
Block dynamics refers to the analysis of forces and motion of a block involved in the problem. Here, understanding Newton鈥檚 Laws of Motion is key. These laws help in describing how the block will move under various forces like tension and friction.
When the block slides, the dynamics include checking how forces work together or against each other to affect motion. The balance of forces -- such as the pull from tension and the resistance from kinetic friction -- determines movement types and speeds.
  • Determine external forces (like tension).
  • Identify resistive forces (like friction).
  • Understand interaction results in net force and acceleration.
This knowledge helps in predicting motion, ensuring you can accurately calculate how the block behaves on its slide.

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Most popular questions from this chapter

A skater with an initial speed of \(7.60 \mathrm{m} / \mathrm{s}\) stops propelling himself and begins to coast across the ice, eventually coming to rest. Air resistance is negligible. (a) The coefficient of kinetic friction between the ice and the skate blades is \(0.100 .\) Find the deceleration caused by kinetic friction. (b) How far will the skater travel before coming to rest?

At an instant when a soccer ball is in contact with the foot of a player kicking it, the horizontal or \(x\) component of the ball's acceleration is \(810 \mathrm{m} / \mathrm{s}^{2}\) and the vertical or \(y\) component of its acceleration is \(1100 \mathrm{m} / \mathrm{s}^{2} .\) The ball's mass is \(0.43 \mathrm{kg} .\) What is the magnitude of the net force acting on the soccer ball at this instant?

A rescue helicopter is lifting a man (weight \(=822 \mathrm{N}\) ) from a capsized boat by means of a cable and harness. (a) What is the tension in the cable when the man is given an initial upward acceleration of \(1.10 \mathrm{m} / \mathrm{s}^{2} ?\) (b) What is the tension during the remainder of the rescue when he is pulled upward at a constant velocity?

A bicyclist is coasting straight down a hill at a constant speed. The combined mass of the rider and bicycle is \(80.0 \mathrm{kg}\), and the hill is inclined at \(15.0^{\circ}\) with respect to the horizontal. Air resistance opposes the motion of the cyclist. Later, the bicyclist climbs the same hill at the same constant speed. How much force (directed parallel to the hill) must be applied to the bicycle in order for the bicyclist to climb the hill?

Two blocks are sliding to the right across a horizontal surface, as the drawing shows. In Case A the mass of each block is \(3.0 \mathrm{kg} .\) In Case \(\mathrm{B}\) the mass of block 1 (the block behind) is \(6.0 \mathrm{kg}\), and the mass of block 2 is \(3.0 \mathrm{kg} .\) No frictional force acts on block 1 in either \(\mathrm{Case}\) A or \(\mathrm{Case} \mathrm{B}\). However, a kinetic frictional force of \(5.8 \mathrm{N}\) does act on block 2 in both cases and opposes the motion. For both Case A and Case B determine (a) the magnitude of the forces with which the blocks push against each other and (b) the magnitude of the acceleration of the blocks.
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