Problem 59
Question
\(\bullet\) While a roofer is working on a roof that slants at \(36^{\circ}\) above the horizontal, he accidentally nudges his 85.0 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 \(\mathrm{N} ?\)
Step-by-Step Solution
Verified Answer
The toolbox will be moving at 5.24 m/s when it reaches the edge of the roof.
1Step 1: Define the Problem and Known Values
The problem involves an inclined plane with a toolbox that starts sliding down due to gravity. We know the mass of the toolbox is 85.0 N (weight), the inclination angle is \(36^{\circ}\), the frictional force is 22.0 N, and the distance to the edge is 4.25 m. Our goal is to find the final speed of the toolbox as it reaches the edge.
2Step 2: Calculate the Component of Gravitational Force Down the Slope
The gravitational force acting down the slope can be found using the component of the weight parallel to the plane: \[ F_{\text{gravity, parallel}} = mg \sin(\theta) \]Where \( m = \frac{85.0}{9.8} \approx 8.67 \text{ kg} \) is the mass and \( \theta = 36^{\circ} \).Let's calculate:\[ F_{\text{gravity, parallel}} = 85.0 \sin(36^{\circ}) \approx 50.0 \text{ N} \]
3Step 3: Determine the Net Force Along the Slope
The net force acting on the toolbox is the difference between the gravitational force component down the slope and the frictional force:\[ F_{\text{net}} = F_{\text{gravity, parallel}} - F_{\text{friction}} \]\[ F_{\text{net}} = 50.0 \text{ N} - 22.0 \text{ N} = 28.0 \text{ N} \]
4Step 4: Calculate the Acceleration of the Toolbox
Using Newton's second law, \( F = ma \), we can solve for the acceleration \(a\):\[ a = \frac{F_{\text{net}}}{m} = \frac{28.0 \text{ N}}{8.67 \text{ kg}} \approx 3.23 \text{ m/s}^2 \]
5Step 5: Use Kinematics to Find the Final Velocity
We use the kinematic equation to find the final velocity \( v \) given initial velocity \( u=0 \), acceleration \( a \), and distance \( s \):\[ v^2 = u^2 + 2as \]Substitute the known values:\[ v^2 = 0 + 2 \times 3.23 \times 4.25 \]\[ v^2 = 27.455 \]\[ v = \sqrt{27.455} \approx 5.24 \text{ m/s} \]
6Step 6: Conclusion
The toolbox will be moving at approximately 5.24 m/s just as it reaches the edge of the roof.
Key Concepts
Kinetic FrictionNewton's Second LawGravitational Force ComponentsKinematics
Kinetic Friction
Kinetic friction is the force that opposes the motion of two surfaces sliding against each other. In the context of our problem, this frictional force acts on the toolbox as it slides down the inclined roof.
It is important to distinguish between kinetic and static friction; the latter acts to prevent motion, while kinetic friction acts during motion. The kinetic friction force is given as 22.0 N, directly opposing the toolbox's motion.
This resistive force plays a crucial role in determining how quickly the toolbox accelerates. The magnitude of kinetic friction depends on the materials involved and the normal force, but in this problem, it is given precisely.
Understanding kinetic friction helps us realize why the toolbox doesn't accelerate as fast as it would if it were frictionless. As a result, it must overcome this force to slide, thereby reducing its net acceleration. By calculating net force, we incorporate the effect of this friction, making our calculations realistic and accurate.
It is important to distinguish between kinetic and static friction; the latter acts to prevent motion, while kinetic friction acts during motion. The kinetic friction force is given as 22.0 N, directly opposing the toolbox's motion.
This resistive force plays a crucial role in determining how quickly the toolbox accelerates. The magnitude of kinetic friction depends on the materials involved and the normal force, but in this problem, it is given precisely.
Understanding kinetic friction helps us realize why the toolbox doesn't accelerate as fast as it would if it were frictionless. As a result, it must overcome this force to slide, thereby reducing its net acceleration. By calculating net force, we incorporate the effect of this friction, making our calculations realistic and accurate.
Newton's Second Law
Newton's second law of motion is a fundamental principle used to relate the forces acting on an object to its motion. It states that the force on an object is equal to its mass times its acceleration, described mathematically as:
First, we need to determine the gravitational force pulling the toolbox down the roof. From this, we subtract the kinetic friction force. This difference gives us the net force, which is responsible for the toolbox's acceleration down the slope.
With the net force and the mass, Newton's second law allows us to find the acceleration. This step is crucial because knowing the acceleration helps us predict how the velocity of the toolbox changes as it slides. In this problem, it enables us to progress to finding the final velocity.
- \( F = ma \)
First, we need to determine the gravitational force pulling the toolbox down the roof. From this, we subtract the kinetic friction force. This difference gives us the net force, which is responsible for the toolbox's acceleration down the slope.
With the net force and the mass, Newton's second law allows us to find the acceleration. This step is crucial because knowing the acceleration helps us predict how the velocity of the toolbox changes as it slides. In this problem, it enables us to progress to finding the final velocity.
Gravitational Force Components
When objects are on inclined planes, the gravitational force can be split into components: one perpendicular to the inclined surface and another parallel. Understanding these components is key to solving the problem of motion on a slope.
The component parallel to the incline is calculated using:
In our scenario, the parallel component propels the toolbox down the roof, becoming a vital part of finding the net force. As gravity pulls it down, we need to counterbalance it with the kinetic friction to find the actual force responsible for motion.
This gravitational force’s parallel component is what partially "fights" friction, allowing it to move. Understanding how to break down forces into components helps us analyze similar problems systematically.
The component parallel to the incline is calculated using:
- \( F_{\text{gravity, parallel}} = mg \sin(\theta) \)
In our scenario, the parallel component propels the toolbox down the roof, becoming a vital part of finding the net force. As gravity pulls it down, we need to counterbalance it with the kinetic friction to find the actual force responsible for motion.
This gravitational force’s parallel component is what partially "fights" friction, allowing it to move. Understanding how to break down forces into components helps us analyze similar problems systematically.
Kinematics
Kinematics involves analyzing the motion of objects without considering the forces that cause this motion. It allows us to relate various physical quantities, such as velocity, acceleration, and displacement.
In the case of the toolbox, we apply kinematic equations to calculate its final velocity after sliding down the inclined roof. Given the initial velocity (starting from rest), acceleration, and distance, the kinematic equation we use is:
By plugging the known values into this equation, we find the speed of the toolbox at the roof's edge. This process illustrates how kinematics enables us to predict motion outcomes in various scenarios, providing clear insights into the dynamic behavior of objects.
In the case of the toolbox, we apply kinematic equations to calculate its final velocity after sliding down the inclined roof. Given the initial velocity (starting from rest), acceleration, and distance, the kinematic equation we use is:
- \( v^2 = u^2 + 2as \)
By plugging the known values into this equation, we find the speed of the toolbox at the roof's edge. This process illustrates how kinematics enables us to predict motion outcomes in various scenarios, providing clear insights into the dynamic behavior of objects.
Other exercises in this chapter
Problem 57
\(\bullet\) A 12.0 g plastic ball is dropped from a height of 2.50 \(\mathrm{m}\) and is moving at 3.20 \(\mathrm{m} / \mathrm{s}\) just before it hits the floo
View solution Problem 58
\(\bullet\) \(\bullet\) You and three friends stand at the corners of a square whose sides are 8.0 \(\mathrm{m}\) long in the mid- dle of the gym floor, as show
View solution Problem 61
\(\bullet\) \(\bullet\) A loaded 375 kg toboggan is traveling on smooth hori- zontal snow at 4.5 \(\mathrm{m} / \mathrm{s}\) when it suddenly comes to a rough r
View solution Problem 62
\(\bullet\) \(\bullet\) A 62.0 kg skier is moving at 6.50 \(\mathrm{m} / \mathrm{s}\) on a frictionless, horizontal snow-covered plateau when she encounters a r
View solution