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Chapter 10: Problem 54

Bundle 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 will slide approximately 346 meters up the incline.

Step by step solution

01

Identify the Given Data

Mass of the bundle, m = 4.0 kg; Initial kinetic energy, KE = 128 J; Incline angle, \( \theta = 30^{\circ} \); Coefficient of kinetic friction, \( \mu_k = 0.30 \).
02

Calculate the Normal Force

Using \( N = mg \cos \theta \), where g is the acceleration due to gravity (9.8 m/s虏), \( N = 4.0 \text{ kg} \times 9.8 \text{ m/s}^2 \times \cos 30^{\circ} \approx 33.9 \text{ N} \).
03

Calculate the Frictional Force

Using \( f_k = \mu_k N \), \( f_k = 0.30 \times 33.9 \text{ N} \approx 10.17 \text{ N} \).
04

Calculate the Work Done by Friction

The work done by friction (W_f) is \( f_k \times d \cos 180^{\circ} \). Since force and distance are in opposite directions, \( W_f = -f_k \times d \).
05

Calculate the Work Done Against Gravity

The work done against gravity (W_g) is \( mgd \sin \theta \).
06

Apply Conservation of Energy

Initial kinetic energy is used to do work against friction and gravity: \( KE = W_f + W_g \). Thus, \( 128 \text{ J} = -10.17d + 4.0 \text{ kg} \times 9.8 \text{ m/s}^2 \times d \sin 30^{\circ} \).
07

Solve for the Distance

Combine and solve for d: \( 128 = -10.17d + 19.6d \, \sin 30^{\circ} = -10.17d + 19.6d \, /2 = 128 \rightarrow d (19.6 / 2 - 10.17) = 128 \rightarrow d (9.8 - 10.17) = 128 \rightarrow d \times -0.37 \approx 128 \rightarrow d \approx -128 / -0.37 \approx 346 \text{ m} \).

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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 the energy that an object possesses due to its motion. It is given by the equation: \[ KE = \frac{1}{2} mv^2 \]where **m** is the mass and **v** is the velocity. In the given problem, our starting point is the initial kinetic energy of the bundle, which is 128 Joules. This energy will be transformed as the bundle moves up the inclined plane.

Understanding kinetic energy helps us predict how far the bundle will slide. As the bundle travels up, this kinetic energy will be used against forces like friction and gravity, reducing its speed until it stops.
coefficient of friction
The coefficient of friction is a measure of how much frictional force exists between two surfaces. It is usually denoted by the symbol \( \mu \). In this problem, we deal with the coefficient of kinetic friction, which comes into play when objects are sliding against each other.

The coefficient given is \( \mu_k = 0.30 \), meaning that 30% of the normal force will be turned into kinetic friction. This frictional force will work against the motion, reducing the kinetic energy of the bundle as it moves up the incline.

Knowing the coefficient of friction helps us calculate the frictional force using the formula: \[ f_k = \mu_k N \]where **N** is the normal force.
conservation of energy
The principle of conservation of energy states that energy cannot be created or destroyed, only transformed from one form to another. In our scenario, the bundle's initial kinetic energy will be converted into work done against friction and gravity.

Since no energy is lost, the initial kinetic energy equals the sum of the work done by friction and the work done against gravity:
\[ KE = W_f + W_g \]This principle allows us to set up our equation to solve for the distance the bundle will slide. By knowing how much energy is used up in friction and in overcoming gravitational pull, we can determine how far the bundle travels before coming to a stop.
work done by friction
Work done by friction can be calculated using the frictional force and the distance over which it acts. The formula is: \[ W_f = f_k d \]where **f_k** is the kinetic friction force, and **d** is the distance. In our problem, friction opposes motion, so the work done is negative:
\[ W_f = -f_k d \]

The friction here interacts with the surface of the inclined plane and the bundle, gradually using up the kinetic energy. Since we calculated the frictional force as 10.17 N, we can account for how much energy is consumed by this force over the distance the bundle slides.
normal force
The normal force is the perpendicular force exerted by a surface on an object resting on it. For an inclined plane, it can be calculated using the formula:
\[ N = mg \cos \theta \]
where **m** is the mass, **g** is the acceleration due to gravity, and \( \cos \theta \) is the cosine of the incline angle.

In the given exercise, this translates to:
\[ N = 4.0 kg \times 9.8 m/s^2 \times \cos 30^{\circ} = 33.9 N \]
This normal force is essential for calculating the frictional force, as it directly affects how much friction is present between the bundle and the inclined plane.
work done against gravity
Work done against gravity refers to the energy required to move an object against the gravitational pull. It can be calculated using the formula:
\[ W_g = mgd \sin \theta \]
where **m** is the mass, **g** is the acceleration due to gravity, **d** is the distance, and \( \sin \theta \) is the sine of the angle of the incline.

In the current problem, the bundle's motion up the incline requires work to be done against gravity. As the bundle moves, this work uses up some of its kinetic energy, limiting how far it can slide. This balance between kinetic energy and work against gravity is crucial for determination of the distance the bundle will travel.

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Most popular questions from this chapter

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