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Chapter 8: Problem 60

A \(4.0 \mathrm{~kg}\) bundle starts up a \(30^{\circ}\) incline with \(128 \mathrm{~J}\) of kinetic energy. How far will it slide up the incline if the coefficient of kinetic friction between bundle and incline is \(0.30 ?\)

Short Answer

Expert verified
The bundle slides 4.3 meters up the incline.

Step by step solution

01

Calculate the Initial Velocity

To determine how far the bundle will slide, we first need to calculate its initial velocity. The kinetic energy is given by \( KE = \frac{1}{2}mv^2 \). Plug in the values: \( 128 = \frac{1}{2} \times 4.0 \times v^2 \). Solve for \( v \): \( v = \sqrt{\frac{256}{4}} = 8 \, \text{m/s} \).
02

Determine Forces Acting on the Bundle

Identify the forces acting on the bundle as it slides up the incline. These include gravity, normal force, and kinetic friction. The force due to gravity acting parallel to the incline is \( mg \sin \theta \), and the frictional force is \( f_k = \mu_k \cdot N \), where \( N = mg \cos \theta \). Here, \( \theta = 30^{\circ}, \mu_k = 0.30, m = 4.0 \) kg.
03

Calculate the Frictional Force

Calculate the normal force: \( N = mg \cos \theta = 4.0 \times 9.8 \times \cos 30^{\circ} = 33.94 \) N. Then, calculate the frictional force: \( f_k = \mu_k \cdot N = 0.30 \times 33.94 = 10.18 \) N.
04

Find the Net Force Along the Incline

Calculate the gravitational component along the incline: \( F_{gravity} = mg \sin \theta = 4.0 \times 9.8 \times \sin 30^{\circ} = 19.6 \) N. The total retarding force is the sum of frictional and gravitational forces: \( f_{total} = f_k + F_{gravity} = 10.18 + 19.6 = 29.78 \) N.
05

Calculate Acceleration

Using Newton's second law, \( F = ma \), find the acceleration: \( a = \frac{f_{total}}{m} = \frac{29.78}{4.0} = 7.45 \, \text{m/s}^2 \). The negative sign indicates deceleration.
06

Calculate the Distance up the Incline

Use the kinematic equation \( v^2 = u^2 + 2as \) to find the distance \( s \). Here, final velocity \( v = 0 \) (at the highest point), initial velocity \( u = 8 \) m/s, and acceleration \( a = -7.45 \) m/s虏: \( 0 = 8^2 + 2(-7.45)s \). Solve for \( s \): \( s = \frac{64}{14.9} = 4.3 \) m.

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Key Concepts

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

Kinetic Energy
Kinetic energy is the energy that an object possesses because of its motion.
This energy is given by the formula \( KE = \frac{1}{2}mv^2 \), where \( m \) is the mass of the object and \( v \) is its velocity.
In the problem, the bundle starts with a kinetic energy of \( 128 \) Joules.
Understanding kinetic energy is essential because it helps us predict how far the object can travel based on its current energy level.
  • A higher kinetic energy means the object can perform more work before stopping.
  • Conversely, if kinetic energy reduces, the object covers less distance.
By rearranging the kinetic energy equation, we deduced the initial velocity of the bundle as it starts moving up the incline.
This initial velocity is crucial for further calculations when determining how far the bundle travels before coming to a stop.
Forces on an Incline
When an object is on an incline, it experiences various forces.
These include the gravitational force, normal force, and any other applied forces like friction.
In this exercise, the incline creates an angle of \( 30^{\circ} \), affecting the net forces on the bundle.
  • The gravitational force acts downward, but it can be broken into components parallel and perpendicular to the incline.
  • The parallel component pulls the object down the slope, calculated as \( mg \sin \theta \).
  • The perpendicular component, \( mg \cos \theta \), affects the normal force.
Understanding these components is valuable as they help us analyze motion on slopes and predict whether the object moves or stays still.
This component-based approach simplifies the problem-solving process, especially when dealing with inclined planes.
Kinetic Friction
Kinetic friction acts as a force that opposes the motion of an object sliding across a surface.
It is vital in this problem because it reduces the bundle's kinetic energy as it ascends the incline.
  • The force of kinetic friction, \( f_k \), is calculated using the formula \( f_k = \mu_k \cdot N \), where \( \mu_k \) is the coefficient of kinetic friction, and \( N \) is the normal force.
  • This friction coefficient \( \mu_k \) is a unitless number that represents how rough or smooth two surfaces interact.
  • A higher \( \mu_k \) indicates more friction, requiring more force to slide an object.
In this case, \( \mu_k \) is given as \( 0.30 \).
Therefore, knowing the forces acting on the bundle and the coefficient of kinetic friction allows us to calculate how quickly energy is lost due to friction, affecting how far the bundle slides up the incline.
Newton's Second Law
Newton's Second Law provides a framework for understanding the relationship between an object's mass, acceleration, and applied force.
The law is expressed as \( F = ma \), meaning the net force on an object is equal to the mass of the object multiplied by its acceleration.
  • This law helps us identify how forces like friction and gravity alter the bundle's motion as it moves up the incline.
  • By determining the total retarding force from both gravity and friction, we can find the net force acting against the bundle.
  • Using the mass of the bundle, we calculate the acceleration, which in this case, results in deceleration.
The application of Newton's Second Law is crucial here.
It connects the forces with the object's motion, thus allowing us to use kinematic equations to solve for the distance the bundle travels up the hill before stopping.

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Most popular questions from this chapter

A collie drags its bed box across a floor by applying a horizontal force of \(8.0 \mathrm{~N}\). The kinetic frictional force acting on the box has magnitude \(5.0 \mathrm{~N}\). As the box is dragged through \(0.70 \mathrm{~m}\) along the way, what are (a) the work done by the collie's applied force and (b) the increase in thermal energy of the bed and floor?

A rope is used to pull a \(3.57 \mathrm{~kg}\) block at constant speed \(4.06 \mathrm{~m}\) along a horizontal floor. The force on the block from the rope is \(7.68 \mathrm{~N}\) and directed \(15.0^{\circ}\) above the horizontal. What are (a) the work done by the rope's force, (b) the increase in thermal energy of the block-floor system, and (c) the coefficient of kinetic friction between the block and floor?

If a \(70 \mathrm{~kg}\) baseball player steals home by sliding into the plate with an initial speed of \(10 \mathrm{~m} / \mathrm{s}\) just as he hits the ground, (a) what is the decrease in the player's kinetic energy and (b) what is the increase in the thermal energy of his body and the ground along which he slides?

The surface of the continental United States has an area of about \(8 \times 10^{6} \mathrm{~km}^{2}\) and an average elevation of about \(500 \mathrm{~m}\) (above sea level). The average yearly rainfall is \(75 \mathrm{~cm}\). The fraction of this rainwater that returns to the atmosphere by evaporation is \(\frac{2}{3}\); the rest eventually flows into the ocean. If the decrease in gravitational potential energy of the water-Earth system associated with that flow could be fully converted to electrical energy, what would be the average power? (The mass of \(1 \mathrm{~m}^{3}\) of water is \(1000 \mathrm{~kg}\).)

A rope is used to pull a \(3.57 \mathrm{~kg}\) block at constant speed \(4.06 \mathrm{~m}\) along a horizontal floor. The force on the block from the rope is \(7.68 \mathrm{~N}\) and directed \(15.0^{\circ}\) above the horizontal. What are (a) the work done by the rope's force, (b) the increase in thermal energy of the block-floor system, and (c) the coefficient of kinetic friction between the block and floor?
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