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Chapter 5: Problem 65

A box of mass \(12.0 \mathrm{~kg}\) sits at rest on a horizontal surface. The coefficient of kinetic friction between the surface and the box is \(0.300 .\) The box is initially at rest, and then a constant force of magnitude \(F\) and direction \(37.0^{\circ}\) below the horizontal is applied to the box; the box slides along the surface. (a) What is \(F\) if the box has a speed of \(6.00 \mathrm{~m} / \mathrm{s}\) after traveling a distance of \(8.00 \mathrm{~m} ?\) (b) What is \(F\) if the surface is friction less and all the other quantities are the same? (c) What is \(F\) if all the quantities are the same as in part (a) but the force applied to the box is horizontal?

Short Answer

Expert verified
The magnitudes of force required are: (a) 53.28N, (b) 18N, (c) 35.28N.

Step by step solution

01

Calculate the friction force

The friction force can be calculated using the formula \(F_{f}=\mu_{k} m g\), where \(m=12.0 \mathrm{~kg}\) is the mass of the box, \(g=9.8 \mathrm{~m}/\mathrm{s}^{2}\) is the acceleration due to gravity and \(\mu_{k}=0.3\) is the coefficient of kinetic friction. After calculation, we get: \(F_{f}=0.3 \times 12.0 \times 9.8 = 35.28 \mathrm{N}\)
02

Solve for force with angle below the horizontal

In the horizontal direction, the friction force must be equals to the horizontal component of \(F\). The horizontal component of \(F\) can be calculated using the formula \(F \cos(\theta)\), where \(\theta = 37.0^{\circ}\). So, we have: \(F_{f} = F \cos(\theta)\). From this, we can calculate \(F\).
03

Solve for acceleration

The acceleration a can be obtained from the kinematic equation \(v^{2}=u^{2}+2 a s\), where \(v=6.00 \mathrm{~m}/\mathrm{s}\) is the final velocity, \(u=0 \mathrm{~m}/\mathrm{s}\) is the initial velocity and \(s=8.00 \mathrm{~m}\) is the distance. From calculation, we have \(a = v^{2} / 2 s = 1.5 \mathrm{~m}/\mathrm{s}^{2}\)
04

Calculate F for parts (a)

Based on Newton's second law, we can write \(F = m a + F_{f}\), where, \(m\) is the mass of the box, \(a\) is the acceleration and \(F_{f}\) is the friction force. Substituting the values, we get \(F = 12.0 \times 1.5 + 35.28 = 53.28 \mathrm{N}\)
05

Calculate F for part (b)

Since the surface is frictionless, the friction force becomes zero in this case. Thus, \(F = m a = 12.0 \times 1.5 = 18 \mathrm{N}\)
06

Calculate F for part (c)

In this case, the force applied to the box is horizontal; therefore, \(F\) should just balance out the friction force. Therefore, \(F = F_{f} = 35.28 \mathrm{N}\)

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Key Concepts

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

Kinetic Friction
When an object slides over a surface, the resistance it experiences is due to friction. There are two main types of friction: static and kinetic. Here, we focus on kinetic friction, which acts when the object is already in motion.
Kinetic friction is influenced by two main factors:
  • The coefficient of kinetic friction, denoted as \(\mu_k\), a value that depends on the materials in contact. In this exercise, \(\mu_k = 0.300\).
  • The normal force, which is typically the gravitational force normal (or perpendicular) to the surface.
The force of kinetic friction, \(F_f\), can be calculated using the formula \(F_f = \mu_k \cdot m \cdot g\), where \(m\) is the object's mass and \(g\) is the acceleration due to gravity (\(9.8\, \mathrm{m/s}^2\)). This force tries to slow down or stop the movement of the object. Understanding how kinetic friction interacts with other forces is essential for solving problems like this one.
Newton's Second Law
Newton's Second Law of Motion is a fundamental principle that describes the relationship between the force exerted on an object, its mass, and its acceleration. It is usually expressed with the formula:\[ F = m \cdot a \]where \(F\) is the net force applied, \(m\) is the mass, and \(a\) is the acceleration.
This law is incredibly important in physics because it tells us that the force needed to move an object is directly proportional to the object's mass and how quickly we want to change its velocity.
In problems involving friction, such as this one, Newton's Second Law helps balance all forces acting on the object. This includes external forces like the push or pull and frictional forces that oppose motion.
By applying this law, we can determine the net force required to move an object at a certain speed, taking into account factors such as weight and surface interactions. Remember, zero net force means the object moves at constant velocity, or stays at rest, if it was initially stationary.
Kinematics
Kinematics is the branch of physics that deals with the motion of objects, excluding the influence of forces and masses on the objects under study. In kinematics, we explore how objects move using formulas that relate different aspects of their motion, like velocity, acceleration, distance, and time.
One of the basic equations used in kinematics when dealing with problems like this is the formula:\[ v^2 = u^2 + 2as \]This formula connects:
  • \(v\), the final velocity, which is the speed of the object after traveling a distance.
  • \(u\), the initial velocity, which in this case is 0, since the box starts from rest.
  • \(a\), the acceleration, which describes how quickly the object changes its speed.
  • \(s\), the distance over which the object travels, given as 8.00 m in this exercise.
By organizing and manipulating these variables, kinematics allows us to predict or understand the motion of the object over time. It works hand-in-hand with dynamics, which incorporates forces like kinetic friction to fully describe the scenario.

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Most popular questions from this chapter

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?

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?

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?

A rocket of initial mass \(125 \mathrm{~kg}\) (including all the contents) has an engine that produces a constant vertical force (the thrust) of \(1720 \mathrm{~N}\). Inside this rocket, a \(15.5 \mathrm{~N}\) electric power supply rests on the floor. (a) Find the initial acceleration of the rocket. (b) When the rocket initially accelerates, how hard does the floor push on the power supply? (Hint: Start with a free-body diagram for the power supply.)

A racetrack curve has radius \(90.0 \mathrm{~m}\) and is banked at an angle of \(18.0^{\circ} .\) The coefficient of static friction between the tires and the roadway is \(0.400 .\) A race car with mass \(1200 \mathrm{~kg}\) rounds the curve with the maximum speed to avoid skidding. (a) As the car rounds the curve, what is the normal force exerted on it by the road? What are the car's (b) radial acceleration and (c) speed?
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