Warning: foreach() argument must be of type array|object, bool given in /var/www/html/web/app/themes/studypress-core-theme/template-parts/header/mobile-offcanvas.php on line 20

The coefficient of static friction between a block and a horizontal floor is \(0.35,\) while the coefficient of kinetic friction is \(0.22 .\) The mass of the block is \(4.6 \mathrm{kg}\) and it is initially at rest. (a) What is the minimum horizontal applied force required to make the block start to slide? (b) Once the block is sliding, if you keep pushing on it with the same minimum starting force as in part (a), does the block move with constant velocity or does it accelerate? (c) If it moves with constant velocity, what is its velocity? If it accelerates, what is its acceleration?

Short Answer

Expert verified
(a) 15.778 N; (b) It accelerates; (c) Acceleration is 1.274 m/s².

Step by step solution

01

Understanding the Problem

We are given coefficients of static and kinetic friction, the mass of the block, and its state of rest initially. We need to find the minimum force to start the block moving and analyze its motion once it starts.
02

Calculating the Normal Force

For a block resting on a horizontal surface, the normal force \( N \) is equal to the gravitational force. Thus, \( N = mg \). Given \( m = 4.6 \) kg and \( g = 9.8 \) m/s², we calculate:\[ N = 4.6 \, \text{kg} \times 9.8 \, \text{m/s}^2 = 45.08 \, \text{N} \]
03

Applying Static Friction Formula

The force needed to start the block moving is determined by the static friction. Use the formula \( f_s = \mu_s N \), where \( \mu_s = 0.35 \). Calculate:\[ f_s = 0.35 \times 45.08 \, \text{N} = 15.778 \, \text{N} \]
04

Start the Block Moving

The minimum force required to overcome static friction and start the block sliding is 15.778 N.
05

Analyzing Motion After Surpassing Static Friction

With the block now sliding, kinetic friction applies. Calculate the force of kinetic friction: \( f_k = \mu_k N \) with \( \mu_k = 0.22 \):\[ f_k = 0.22 \times 45.08 \, \text{N} = 9.9176 \, \text{N} \]
06

Determining the Motion with the Starting Force

The applied force of 15.778 N exceeds the kinetic friction force of 9.9176 N, leaving a net force that causes acceleration. Calculate the net force:\[ F_\text{net} = 15.778 \, \text{N} - 9.9176 \, \text{N} = 5.8604 \, \text{N} \]
07

Calculating Acceleration

Use Newton's second law \( F = ma \) to find the acceleration \( a \) of the block:\[ a = \frac{F_\text{net}}{m} = \frac{5.8604 \, \text{N}}{4.6 \, \text{kg}} = 1.274 \, \text{m/s}^2 \]

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with Vaia!

Key Concepts

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

Static Friction
Static friction is the force that counteracts the initiation of movement. It's why a block doesn’t slide when resting on a surface until a certain force threshold is surpassed. This frictional force depends on the surfaces in contact and is characterized by the coefficient of static friction, represented by \( \mu_s \). Static friction acts to resist motion up to a maximum value.
In this exercise, we calculate it using the formula: \( f_s = \mu_s N \). Here the normal force \( N \) is equivalent to the block’s weight due to gravity, calculated by \( N = mg \). For a 4.6 kg block, the force needed to start it moving (i.e., to overcome static friction) is 15.778 N.
To achieve this, the applied force must reach or exceed this static friction level. Once surpassed, static friction no longer holds the block in place.
Kinetic Friction
Once the block is in motion, the force that opposes its movement changes from static to kinetic friction. This friction is often less than static friction, which explains why less force is needed to continue moving the block than to start it moving. The coefficient of kinetic friction \( \mu_k \) characterizes this friction and is used alongside the normal force \( N \) to find the force of kinetic friction \( f_k = \mu_k N \).
In our example, with \( \mu_k = 0.22 \) and a normal force of 45.08 N, kinetic friction calculates to 9.9176 N. The block continues to experience this resistance as long as it slides. Importantly, unlike static friction, kinetic friction doesn’t change with the speed of the block.
If the applied force were equal to kinetic friction, the block would move at a constant velocity. However, since the applied force here exceeds kinetic friction, it leads to acceleration.
Newton's Laws of Motion
Newton's laws are fundamental in understanding how forces affect motion. Specifically, his second law is essential here: the acceleration \( a \) of an object is directly proportional to the net force, \( F_{net} \), acting on it and inversely proportional to its mass \( m \), expressed as \( F = ma \).
By analyzing the forces acting on the block, including both applied and frictional forces, we determine the net force. This net force then dictates whether and how the block accelerates.
In this exercise, after applying the initial force, the block experiences a net force greater than zero due to the difference between the applied force and the kinetic friction. Thus, according to Newton’s second law, the block doesn't move with constant velocity; it accelerates, confirming the law’s predictions.
Acceleration
Acceleration is the rate at which the velocity of an object changes over time due to applied forces. In the context of this exercise, after overcoming static friction, the block experiences acceleration.
Once the block starts sliding, the force continuing to act (15.778 N) is more than what is needed to merely oppose kinetic friction (9.9176 N). This difference results in a net force which accelerates the block according to Newton's second law.
We calculate its acceleration at 1.274 \( \text{m/s}^2 \), using the formula \( a = F_{net}/m \). This acceleration signifies a continuous increase in velocity, assuming the applied force stays constant and does not light up.

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

Mechanical advantage is the ratio of the force required without the use of a simple machine to that needed when using the simple machine. Compare the force to lift an object with that needed to slide the same object up a frictionless incline and show that the mechanical advantage of the inclined plane is the length of the incline divided by the height of the incline \((d / h\) in Fig. \(4.25).\)
In a movie, a stuntman places himself on the front of a truck as the truck accelerates. The coefficient of friction between the stuntman and the truck is \(0.65 .\) The stuntman is not standing on anything but can "stick" to the front of the truck as long as the truck continues to accelerate. What minimum forward acceleration will keep the stuntman on the front of the truck?
A truck is towing a 1000 -kg car at a constant speed up a hill that makes an angle of \(\alpha=5.0^{\circ}\) with respect to the horizontal. A rope is attached from the truck to the car at an angle of \(\beta=10.0^{\circ}\) with respect to horizontal. Ignore any friction in this problem. (a) Draw an FBD showing all the forces on the car. Indicate the angle that each force makes with either the vertical or horizontal direction. (b) What is the tension in the rope?
In Fig. 4.15 an astronaut is playing shuffleboard on Earth. The puck has a mass of \(2.0 \mathrm{kg} .\) Between the board and puck the coefficient of static friction is 0.35 and of kinetic friction is \(0.25 .\) (a) If she pushes the puck with a force of \(5.0 \mathrm{N}\) in the forward direction, does the puck move? (b) As she is pushing, she trips and the force in the forward direction suddenly becomes \(7.5 \mathrm{N} .\) Does the puck move? (c) If so, what is the acceleration of the puck along the board if she maintains contact between puck and stick as she regains her footing while pushing steadily with a force of \(6.0 \mathrm{N}\) on the puck? (d) She carries her game to the Moon and again pushes a moving puck with a force of \(6.0 \mathrm{N}\) forward. Is the acceleration of the puck during contact more, the same, or less than on Earth? Explain. (tutorial: rough table)
The vertical component of the acceleration of a sailplane is zero when the air pushes up against its wings with a force of \(3.0 \mathrm{kN}\). (a) Assuming that the only forces on the sailplane are that due to gravity and that due to the air pushing against its wings, what is the gravitational force on the Earth due to the sailplane? (b) If the wing stalls and the upward force decreases to \(2.0 \mathrm{kN}\) what is the acceleration of the sailplane?
See all solutions

Recommended explanations on Physics Textbooks

View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.

Sign-up for free