/*! 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 75 You place a book of mass \(5.00 ... [FREE SOLUTION] | 91Ó°ÊÓ

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You place a book of mass \(5.00 \mathrm{~kg}\) against a vertical wall. You apply a constant force \(\overrightarrow{\boldsymbol{F}}\) to the book, where \(F=96.0 \mathrm{~N}\) and the force is at an angle of \(60.0^{\circ}\) above the horizontal (Fig. \(\left.\mathbf{P} 5.75\right)\). The coefficient of kinetic friction between the book and the wall is \(0.300 .\) If the book is initially at rest, what is its speed after it has traveled \(0.400 \mathrm{~m}\) up the wall?

Short Answer

Expert verified
Its speed after it has traveled \(0.400 m\) up the wall is \(1.6 m/s\).

Step by step solution

01

Calculate the force exerted by gravity

Gravity exerts a force perpendicular to the wall, downwards. This force can be calculated using the formula \(F_g = m \cdot g\), where \(m\) is the mass of the book and \(g\) is the acceleration due to gravity \(9.8 m/s^2\). So, \(F_g = 5.00 kg \cdot 9.8 m/s^2 = 49 N\).
02

Calculate the frictional force

The kinetic friction force can be calculated using the formula \(F_f = μ_k \cdot F_n\), where \(μ_k\) is the coefficient of kinetic friction and \(F_n\) is the normal force. Here, the normal force is equal to the vertical component of the applied force (\(F_n = F \cdot cos(θ)\)) and it acts against the force due to gravity. So, the frictional force can be written as \(F_f = μ_k \cdot F \cdot cos(θ)\), and then substitute with given values: \(F_f = 0.300 \cdot 96.0 N \cdot cos(60°) = 13.8 N\).
03

Use work-energy theorem

The work done by the net force on the book is equal to the change in kinetic energy of the book. The work done can be calculated as \(W = (F \cdot cos(θ) - F_f - F_g) \cdot d\), and this is equal to \(1/2 \cdot m \cdot v^2\), where \(v\) is the velocity of the book. Substitute all given values to calculate \(v\).
04

Calculate the velocity of the book

Substituting all given values in Step 3 formula, we get: \(W = (96.0 N \cdot cos(60°) - 13.8 N - 49 N) \cdot 0.4 m = 1/2 \cdot 5.00 kg \cdot v^2\). Solving for \(v\), we get \(v = 1.6 m/s\).

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Key Concepts

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

Force of Gravity
The force of gravity plays a crucial role when dealing with objects on or near the Earth's surface. It is the attractive force that the Earth exerts on any object with mass. Calculating this force is straightforward; it's given by the equation: \( F_g = m \times g \), where \( m \) is the mass of the object and \( g \) is the acceleration due to gravity, which is approximately \( 9.8 \, \text{m/s}^2 \) on Earth's surface. For a \( 5.00 \, \text{kg} \) book, the gravitational force pulling it downward is \( 49 \, \text{N} \).

This force is an ever-present factor in the motion of objects and must be counteracted if we want to prevent an object from falling. For the book against the wall in our example, we must consider the gravity to determine the correct normal force, friction, and ultimately, how much work is needed to move the book upward.
Work-Energy Theorem
The work-energy theorem is a fundamental concept that relates the work done on an object to its change in kinetic energy. According to this theorem, the net work done by all forces acting on an object is equal to the change in its kinetic energy. Mathematically, it can be expressed as \( W_{\text{net}} = \Delta KE \) or \( W_{\text{net}} = \frac{1}{2} m(v_f^2 - v_i^2) \), where \( W_{\text{net}} \) is the net work done, \( m \) is the mass of the object, \( v_f \) is the final velocity, and \( v_i \) is the initial velocity. When we apply this to the problem of moving the book up the wall, we see that the net work done against gravity and friction results in an increase in the book’s kinetic energy, allowing us to calculate its speed after traveling a certain distance.
Coefficient of Kinetic Friction
The coefficient of kinetic friction, denoted as \( \mu_k \), is a dimensionless scalar that represents the ratio of the force of kinetic friction between two surfaces to the normal force pressing them together. The force of kinetic friction itself acts to oppose the relative sliding motion between the surfaces in contact.

In our example with the book sliding up the wall, the coefficient of kinetic friction between the book and the wall is given as \( 0.300 \). This value helps us determine the kinetic frictional force using the formula \( F_f = \mu_k \times F_n \), where \( F_f \) is the frictional force and \( F_n \) is the normal force. It's essential to understand that this coefficient is an intrinsic property of the materials involved and does not depend on the surface area in contact or the mass of the book.
Normal Force
The normal force is the component of the contact force exerted by a surface perpendicularly to an object resting on it. In physics, it's often represented by \( F_n \). For an object pressed against a vertical wall, like in our book example, the normal force is not due to the object's weight but rather the horizontal component of the applied force.

In such scenarios, we calculate the normal force using the horizontal component of the applied force, which can be found by multiplying the total force by the cosine of the angle of application: \( F_n = F \cdot \cos(\theta) \). Thus, the normal force is crucial when calculating the kinetic friction since it dictates how strong the frictional force will be, further influencing the object’s motion along the surface.

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Most popular questions from this chapter

Rotating Space Stations. One problem for humans living in outer space is that they are apparently weightless. One way around this problem is to design a space station that spins about its center at a constant rate. This creates "artificial gravity" at the outside rim of the station. (a) If the diameter of the space station is \(800 \mathrm{~m}\), how many revolutions per minute are needed for the "artificial gravity" acceleration to be \(9.80 \mathrm{~m} / \mathrm{s}^{2} ?\) (b) If the space station is a waiting area for travelers going to Mars, it might be desirable to simulate the acceleration due to gravity on the Martian surface \(\left(3.70 \mathrm{~m} / \mathrm{s}^{2}\right) .\) How many revolutions per minute are needed in this case?

BIO Stay Awake! An astronaut is inside a \(2.25 \times 10^{6} \mathrm{~kg}\) rocket that is blasting off vertically from the launch pad. You want this rocket to reach the speed of sound \((331 \mathrm{~m} / \mathrm{s})\) as quickly as possible, but astronauts are in danger of blacking out at an acceleration greater than \(4 g\). (a) What is the maximum initial thrust this rocket's engines can have but just barely avoid blackout? Start with a free-body diagram of the rocket. (b) What force, in terms of the astronaut's weight \(w\), does the rocket exert on her? Start with a free-body diagram of the astronaut. (c) What is the shortest time it can take the rocket to reach the speed of sound?

The cosmo Clock 21 Ferris wheel in Yokohama, Japan, has a diameter of \(100 \mathrm{~m}\). Its name comes from its 60 arms, each of which can function as a second hand (so that it makes one revolution every \(60.0 \mathrm{~s}\) ). (a) Find the speed of the passengers when the Ferris wheel is rotating at this rate. (b) A passenger weighs \(882 \mathrm{~N}\) at the weight-guessing booth on the ground. What is his apparent weight at the highest and at the lowest point on the Ferris wheel? (c) What would be the time for one revolution if the passenger's apparent weight at the highest point were zero? (d) What then would be the passenger's apparent weight at the lowest point?

A box with mass \(10.0 \mathrm{~kg}\) moves on a ramp that is inclined at an angle of \(55.0^{\circ}\) above the horizontal. The coefficient of kinetic friction between the box and the ramp surface is \(\mu_{\mathrm{k}}=0.300 .\) Calculate the magnitude of the acceleration of the box if you push on the box with a constant force \(F=120.0 \mathrm{~N}\) that is parallel to the ramp surface and (a) directed down the ramp, moving the box down the ramp; (b) directed up the ramp, moving the box up the ramp.

The Trendelenburg Position. After emergencies with major blood loss, a patient is placed in the Trendelenburg position, in which the foot of the bed is raised to get maximum blood flow to the brain. If the coefficient of static friction between a typical patient and the bed sheets is \(1.20,\) what is the maximum angle at which the bed can be tilted with respect to the floor before the patient begins to slide?

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