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While a roofer is working on a roof that slants at \(36^{\circ}\) above the horizontal, he accidentally nudges his \(85.0-\mathrm{N}\) toolbox, causing it to start sliding downward, starting from rest. If it starts 4.25 \(\mathrm{m}\) from the lower edge of the roof, how fast will the toolbox be moving just as it reaches the edge of the roof if the kinetic friction force on it is 22.0 \(\mathrm{N} ?\)

Short Answer

Expert verified
5.24 m/s

Step by step solution

01

Identify Forces Acting on the Toolbox

The forces acting on the toolbox are gravity, normal force, and kinetic friction. You'll first need to resolve the gravitational force into components parallel and perpendicular to the roof. The component parallel to the roof is given by: \[ F_{parallel} = mg \sin(\theta) \]where \( m \) is the mass of the toolbox, \( g \) is the acceleration due to gravity \( (9.8 \mathrm{m/s^2}) \), and \( \theta \) is the angle of the roof. The normal force and gravitational components will be used to determine the net force.
02

Calculate Mass of the Toolbox

Since the weight \( W \) of the toolbox is given as 85.0 N, we can find the mass \( m \) using the equation \( W = mg \):\[m = \frac{W}{g} = \frac{85.0}{9.8} \approx 8.67 \text{ kg}\]
03

Determine Gravitational Force Component

Substitute the mass and angle into the equation for \( F_{parallel} \):\[F_{parallel} = (8.67 \text{ kg}) \times (9.8 \text{ m/s}^2) \times \sin(36^\circ)\]Solving this gives:\[ F_{parallel} \approx 8.67 \times 9.8 \times 0.5878 \approx 49.96 \text{ N}\]
04

Calculate Net Force on the Toolbox

To find the net force \( F_{net} \), subtract the frictional force \( F_{friction} = 22.0 \text{ N} \) from the parallel component of the gravitational force:\[F_{net} = F_{parallel} - F_{friction} = 49.96 \text{ N} - 22.0 \text{ N} \approx 27.96 \text{ N}\]
05

Apply Newton's Second Law to Find Acceleration

Use Newton's second law, \( F = ma \), to find the acceleration \( a \):\[a = \frac{F_{net}}{m} = \frac{27.96}{8.67} \approx 3.23 \text{ m/s}^2\]
06

Use Kinematic Equation to Find Final Velocity

Use the kinematic equation \( v^2 = u^2 + 2as \) to find the final velocity \( v \), where \( u = 0 \text{ m/s} \) (initial velocity), \( a = 3.23 \text{ m/s}^2 \), and \( s = 4.25 \text{ m} \):\[v^2 = 0 + 2 \times 3.23 \times 4.25 \]\[v^2 = 27.485\]\[ v \approx \sqrt{27.485} \approx 5.24 \text{ m/s}\]
07

Conclusion: Determine Final Speed at Edge of Roof

The toolbox will be moving at approximately 5.24 m/s when it reaches the edge of the roof.

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Key Concepts

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

Understanding Kinetic Friction
Kinetic friction is a crucial concept when objects move across surfaces. Unlike static friction, which must be overcome to initiate movement, kinetic friction acts on already moving objects, opposing their motion.
In this exercise, the toolbox on the roof experiences kinetic friction as it slides downward. This frictional force is given as 22.0 N. Kinetic friction usually depends on two factors:
  • The coefficient of kinetic friction between the surfaces in contact.
  • The normal force, which is the perpendicular force pressing the two surfaces together.
In the real world, this means smoother surfaces will have less kinetic friction while rougher surfaces generate more frictional resistance. Understanding this force is vital to solving problems like the one we have here, determining how it impacts the net force on the toolbox.
Newton's Second Law and Its Application
Newton's Second Law gives us the framework to understand how forces impact an object's motion. It states that the acceleration of an object is directly proportional to the net force acting on it, and inversely proportional to its mass, described by the equation: \( F = ma \). This law is instrumental in analyzing the motion of the toolbox.
In our scenario, the net force is the gravitational component along the inclined plane minus the kinetic friction.
After calculating the net force, you can use this law to find the acceleration of the toolbox, providing a clear path to understanding its motion.
This concept is foundational in physics, elucidating how various forces interact to dictate motion in everyday scenarios.
Gravitational Force in Action
Gravitational force is the pull that the Earth exerts on objects, pulling them toward its center.
The gravitational force acting on the toolbox can be broken down into components that interact with the inclined plane.
For instance, the gravitational force has a component parallel to the inclined plane, calculated using:\[ F_{parallel} = mg \sin(\theta) \]where \( m \) is mass, \( g \) is the acceleration due to gravity, and \( \theta \) is the incline angle. Additionally, the perpendicular component (not affecting motion along the plane) helps determine the normal force, impacting kinetic friction.
This breakdown helps in understanding and predicting how objects will behave in situations involving slopes or inclines.
Analyzing Motion on an Inclined Plane
An inclined plane is a flat surface tilted at an angle to the horizontal. It changes the way gravitational force acts on an object, affecting both the normal and parallel forces.
In this problem, the toolbox slides down such a plane due to the gravitational force's parallel component, while the normal force provides balance in the perpendicular direction.
The angle of the plane, here at 36 degrees, alters these forces and, thus, the acceleration. By analyzing the forces at play, one can determine the toolbox's motion characteristics as it travels down the slant.
Calculating Final Velocity
The final velocity refers to the speed of the toolbox just as it reaches the roof's edge. It's found using the kinematic equation:\[ v^2 = u^2 + 2as \]where \( u \) is the initial velocity (0 in our case), \( a \) is the acceleration, and \( s \) is the distance traveled.
This formula allows us to calculate how fast the toolbox will be moving at any given point, given the initial speed, distance, and acceleration.
In this scenario, after plugging in the known values, we get the toolbox's final speed as it tumbles off the roof, giving us a clear understanding of its motion dynamics as it descends.

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Most popular questions from this chapter

An empty crate is given an initial push down a ramp, starting with speed \(v_{0,}\) and reaches the bottom with speed \(v\) and kinetic energy \(K .\) Some books are now placed in the crate, so that the total mass is quadrupled. The coefficient of kinetic friction is constant and air resistance is negligible. Starting again with \(v_{0}\) at the top of the ramp, what are the speed and kinetic energy at the bottom? Explain the reasoning behind your answers.

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