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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 approximately 4.3 meters up the incline.

Step by step solution

01

Determine Initial Conditions

The bundle has an initial kinetic energy of 128 J. Its mass is 4.0 kg, and it is sliding up a 30掳 incline. The coefficient of kinetic friction is 0.30.
02

Calculate the Force of Friction

The force of friction can be calculated using the formula: \( F_{f} = \mu \cdot F_{N} \), where \( \mu = 0.30 \) and \( F_{N} = mg \cos{30^{\circ}} \). With \( m = 4.0 \, \text{kg} \) and \( g = 9.8 \, \text{m/s}^2 \), we get:\[ F_{N} = 4.0 \cdot 9.8 \cdot \cos{30^{\circ}} \approx 33.94 \, \text{N} \]\[ F_{f} = 0.30 \cdot 33.94 \approx 10.18 \, \text{N} \]
03

Calculate Work Done Against Friction and Gravity

The forces doing work against the bundle are friction and gravity. The gravitational work component can be calculated using \( mg \sin{30^{\circ}} \). Therefore:Work against gravity per meter is \( 4.0 \cdot 9.8 \cdot \sin{30^{\circ}} = 19.6 \, \text{N} \). Total force against motion = \( 19.6 + 10.18 = 29.78 \, \text{N} \).
04

Set up the Energy Balance Equation

The initial kinetic energy will be used to overcome the work done against friction and gravity. Let \( d \) be the distance traveled, then work done can be expressed as:\[ 29.78 \cdot d = 128 \]
05

Solve for Distance

Rearrange the energy balance equation to solve for distance \( d \):\[ d = \frac{128}{29.78} \approx 4.3 \, \text{m} \]

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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 a type of energy associated with the motion of an object. It depends on the mass of the object and its velocity. The formula to calculate kinetic energy is given by: \( KE = \frac{1}{2}mv^2 \)In this equation, \( KE \) stands for kinetic energy, \( m \) is the mass, and \( v \) represents the velocity of the object.
  • More mass or speed means more kinetic energy.
  • This energy can be transformed into other forms, like potential energy or heat, when work is done.
In the original problem, the bundle starts with a kinetic energy of 128 J as it begins to slide upward an incline. That's because the bundle is moving and thus has energy due to its motion.
Friction
Friction is the force that opposes the motion of objects sliding against each other. It's an important concept because it causes some of the kinetic energy to be lost as heat.
  • Friction depends on the coefficient of friction (\( \mu \)) and the normal force (\( F_N \)) pressing the surfaces together.
  • The force of friction \( F_f \) is calculated by: \( F_f = \mu \cdot F_N \).
In our exercise, the coefficient of kinetic friction is 0.30, meaning 30% of the normal force is acting against the bundle's motion. As the bundle slides up, this frictional force reduces the amount of distance it can travel by using up some of its initial kinetic energy.
Inclined Plane
An inclined plane is a flat surface tilted at an angle, used to help raise or lower objects. When an object moves on an inclined plane, gravity has a component acting down the slope and this affects the motion of the object.
  • The angle of the slope affects both the normal force and the gravitational force acting parallel to the slope.
  • The equation \( F_N = mg \cos{\theta} \) calculates the normal force on an incline.
  • The gravitational force along the plane is \( mg \sin{\theta} \), where \( \theta \) is the angle of the incline.
In the scenario, the incline is set at 30掳, which affects how the forces of gravity and friction interact with the object's motion.
Work and Energy Concepts
The Work and Energy concepts help us understand how energy is transformed or transferred within a system. "Work" refers to the process of energy transfer when a force is applied over a distance. The amount of work done depends on the force exerted and the distance moved in the direction of the force.
  • The work-energy principle states that the work done on an object is equal to the change in kinetic energy of the object.
  • Forces like friction and gravity cause changes in height or motion, converting energy from one form to another.
  • In this problem, the initial kinetic energy was used to do work against friction and gravity as the bundle slides up the incline.
The balance of these energy types and the forces is described by the equation \( 29.78 \cdot d = 128 \), and solving this gives the distance the bundle can slide before it stops.

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Most popular questions from this chapter

A pendulum consists of a \(2.0 \mathrm{~kg}\) stone swinging on a \(4.0 \mathrm{~m}\) string of negligible mass. The stone has a speed of \(8.0 \mathrm{~m} / \mathrm{s}\) when it passes its lowest point. (a) What is the speed when the string is at \(60^{\circ}\) to the vertical? (b) What is the greatest angle with the vertical that the string will reach during the stone's motion? (c) If the potential energy of the pendulum-Earth system is taken to be zero at the stone's lowest point, what is the total mechanical energy of the system?

A locomotive with a power capability of \(1.5 \mathrm{MW}\) can accelerate a train from a speed of \(10 \mathrm{~m} / \mathrm{s}\) to \(25 \mathrm{~m} / \mathrm{s}\) in \(6.0 \mathrm{~min}\). (a) Calculate the mass of the train. Find (b) the speed of the train and (c) the force accelerating the train as functions of time (in seconds) during the \(6.0\) min interval. (d) Find the distance moved by the train during the interval.

A \(75 \mathrm{~g}\) Frisbee is thrown from a point \(1.1 \mathrm{~m}\) above the ground with a speed of \(12 \mathrm{~m} / \mathrm{s}\). When it has reached a height of \(2.1 \mathrm{~m}\), its speed is \(10.5 \mathrm{~m} / \mathrm{s}\). What was the reduction in \(E_{\mathrm{mec}}\) of the Frisbee-Earth system because of air drag?

A block is released from rest at height \(d=40\) \(\mathrm{cm}\) and slides down a frictionless ramp and onto a first plateau, which has length \(d\) and where the coefficient of kinetic friction is \(0.50\). If the block is still moving, it then slides down a second frictionless ramp through height \(d / 2\) and onto a lower plateau, which has length \(d / 2\) and where the coefficient of kinetic friction is again \(0.50 .\) If the block is still moving, it then slides up a frictionless ramp until it (momentarily) stops. Where does the block stop? If its final stop is on a plateau, state which one and give the distance \(L\) from the left edge of that plateau. If the block reaches the ramp, give the height \(H\) above the lower plateau where it momentarily stops.

A worker pushed a \(27 \mathrm{~kg}\) block \(9.2 \mathrm{~m}\) along a level floor at constant speed with a force directed \(32^{\circ}\) below the horizontal. If the coefficient of kinetic friction between block and floor was \(0.20\), what were (a) the work done by the worker's force and (b) the increase in thermal energy of the block- floor system?
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