/*! 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 37 A block whose weight is \(45.0 \... [FREE SOLUTION] | 91Ó°ÊÓ

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A block whose weight is \(45.0 \mathrm{~N}\) rests on a horizontal table. A horizontal force of \(36.0 \mathrm{~N}\) is applied to the block. The coefficients of static and kinetic friction are 0.650 and 0.420 , respectively. Will the block move under the influence of the force, and, if so, what will be the block's acceleration? Explain your reasoning.

Short Answer

Expert verified
Yes, the block will move with an acceleration of approximately 3.72 m/s².

Step by step solution

01

Calculate the Normal Force

The normal force is equal to the weight of the block because it rests on a horizontal table. Thus, \[ N = W = 45.0 \, \text{N} \]
02

Determine the Maximum Static Friction Force

Use the coefficient of static friction to find the maximum static friction force using the formula:\[ f_{s\text{ max}} = \mu_s \cdot N = 0.650 \cdot 45.0 \, \text{N} = 29.25 \, \text{N} \]
03

Compare Applied Force with Static Friction

Compare the applied force with the maximum static friction force:- Applied force: \( 36.0 \, \text{N} \)- Maximum static friction: \( 29.25 \, \text{N} \)Since \( 36.0 \, \text{N} > 29.25 \, \text{N} \), the block will move.
04

Calculate Kinetic Friction Force

Once the block is moving, use the coefficient of kinetic friction to find the kinetic friction force:\[ f_k = \mu_k \cdot N = 0.420 \cdot 45.0 \, \text{N} = 18.9 \, \text{N} \]
05

Determine the Net Force

The net force acting on the block once it starts moving is given by:\[ F_{\text{net}} = F_{\text{applied}} - f_k = 36.0 \, \text{N} - 18.9 \, \text{N} = 17.1 \, \text{N} \]
06

Calculate the Acceleration

Use Newton's second law to calculate the acceleration of the block. Let the mass \( m \) be what's responsible for the given weight so \( m = \frac{W}{g} = \frac{45.0 \, \text{N}}{9.8 \, \text{m/s}^2} = 4.59 \, \text{kg} \). Now find acceleration:\[ a = \frac{F_{\text{net}}}{m} = \frac{17.1 \, \text{N}}{4.59 \, \text{kg}} = 3.72 \, \text{m/s}^2 \]

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Key Concepts

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

Static Friction
Static friction is the force that comes into play when an object is at rest and a force is trying to move it. It is the friction that must be overcome to start moving the object. This type of friction works to resist the initial motion between two surfaces that are in contact. It always acts in the opposite direction to the applied force.
In our problem, the static friction force is determined using the coefficient of static friction, denoted as \(\mu_s\). The maximum static frictional force is calculated as \(f_{s\text{ max}} = \mu_s \cdot N\), where \(N\) is the normal force, which equals the weight of the block.
  • The coefficient of static friction in this exercise is 0.650, so we calculate the maximum static friction as \(29.25 \, \text{N}\).
  • The applied force is \(36.0 \, \text{N}\), which is greater than the maximum static friction force. Thus, the block overcomes static friction and starts to move.
Static friction is crucial in determining whether an object will start to move when a force is applied. If the applied force exceeds the maximum static frictional force, the object will begin to move.
Kinetic Friction
Once an object starts moving, static friction is no longer at play. Instead, kinetic friction becomes the resisting force. Kinetic friction acts in the opposite direction of the moving object and tends to be less than static friction. This is why objects can appear easier to move once they're already in motion.
In the given situation, we use the coefficient of kinetic friction, \(\mu_k\), to find the kinetic frictional force, calculated as \(f_k = \mu_k \cdot N\). The coefficient of kinetic friction is given as 0.420.
  • Thus, the kinetic friction force is calculated as \(18.9 \, \text{N}\).
  • This force is subtracted from the initial applied force to calculate the net force, as kinetic friction continues to resist the motion.
Understanding kinetic friction helps us predict how a moving object will behave and is essential for calculating further motion dynamics, such as acceleration.
Newton's Second Law
Newton's second law of motion provides the foundation for understanding how forces affect the movement of objects. It states that the acceleration of an object is directly proportional to the net force acting upon it and inversely proportional to its mass. This relationship is expressed as \(F = ma\), where \(F\) is the net force, \(m\) is the mass, and \(a\) is the acceleration.
In our example, we first determined the block's mass by using its weight, as weight is the gravitational force acting on the mass. Using the relation \(W = mg\), we derived the mass as \(4.59 \, \text{kg}\).
  • The net force acting on the block, once in motion and considering kinetic friction, was found to be \(17.1 \, \text{N}\).
  • Using Newton's second law, we calculated the block’s acceleration to be \(3.72 \, \text{m/s}^2\).
Newton’s second law helps us determine the acceleration resulting from forces acting on an object, enabling us to predict and analyze motion effectively.

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Most popular questions from this chapter

Consult Interactive LearningWare 4.2 at before beginning this problem. A truck is traveling at a speed of \(25.0 \mathrm{~m} / \mathrm{s}\) along a level road. A crate is resting on the bed of the truck, and the coefficient of static friction between the crate and the truck bed is \(0.650 .\) Determine the shortest distance in which the truck can come to a halt without causing the crate to slip forward relative to the truck.

Concept Questions Two skaters, a man and a woman, are standing on ice. Neglect any friction between the skate blades and the ice. The woman pushes on the man with a certain force that is parallel to the ground. (a) Must the man accelerate under the action of this force? If so, what three factors determine the magnitude and direction of his acceleration? (b) Is there a corresponding force exerted on the woman? If so, where does it originate? Is this force related to the magnitude and direction of the force the woman exerts on the man? If so, how? Problem The mass of the man is \(82 \mathrm{~kg}\) and that of the woman is \(48 \mathrm{~kg} .\) The woman pushes on the man with a force of \(45 \mathrm{~N}\) due east. Determine the acceleration (magnitude and direction) of (a) the man and (b) the woman.

A Mercedes-Benz 300SL \((m=1700 \mathrm{~kg})\) is parked on a road that rises \(15^{\circ}\) above the horizontal. What are the magnitudes of (a) the normal force and (b) the static frictional force that the ground exerts on the tires?

At a time when mining asteroids has become feasible, astronauts have connected a line between their 3500 -kg space tug and a 6200 -kg asteroid. Using their ship's engine, they pull on the asteroid with a force of \(490 \mathrm{~N}\). Initially the tug and the asteroid are at rest, \(450 \mathrm{~m}\) apart. How much time does it take for the ship and the asteroid to meet?

A \(1.14 \times 10^{4} \mathrm{~kg}\) lunar landing craft is about to touch down on the surface of the moon, where the acceleration due to gravity is \(1.60 \mathrm{~m} / \mathrm{s}^{2}\). At an altitude of \(165 \mathrm{~m}\) the craft's downward velocity is \(18.0 \mathrm{~m} / \mathrm{s}\). To slow down the craft, a retrorocket is firing to provide an upward thrust. Assuming the descent is vertical, find the magnitude of the thrust needed to reduce the velocity to zero at the instant when the craft touches the lunar surface.

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