Problem 10 - University Physics with Modern Physics

Question

A 12.0-kg package in a mail-sorting room slides 2.00 m down a chute that is inclined at 53.0\(^\circ\) below the horizontal. The coefficient of kinetic friction between the package and the chute's surface is 0.40. Calculate the work done on the package by (a) friction, (b) gravity, and (c) the normal force. (d) What is the net work done on the package?

Step-by-Step Solution

Answer

(a) -56.58 J, (b) 188.16 J, (c) 0 J, (d) 131.58 J

Step 1: Identify Forces Acting on the Package

The forces acting on the package are gravity, normal force, and friction. Gravity acts downward, normal force is perpendicular to the chute's surface, and friction opposes the motion along the chute.

Step 2: Calculate the Force of Friction

The force of friction can be calculated using the formula: \[ f_k = \mu_k \cdot N \]where \( \mu_k = 0.40 \) is the coefficient of kinetic friction and \( N \) is the normal force. First, calculate \( N \) using:\[ N = mg\cos\theta \]where \( m = 12.0 \text{ kg} \), \( g = 9.8 \text{ m/s}^2 \), and \( \theta = 53.0^\circ \). Compute \( N = 12.0 \times 9.8 \times \cos(53.0^\circ) \approx 70.72 \text{ N} \).Thus, the force of friction is:\[ f_k = 0.40 \times 70.72 \approx 28.29 \text{ N} \]

Step 3: Calculate Work Done by Friction

Use the formula for work: \( W = f d \cos\phi \), where \( f = f_k = 28.29 \text{ N} \), \( d = 2.00 \text{ m} \), and \( \phi = 180^\circ \) because friction opposes motion.Thus, the work done by friction is:\[ W_{\text{friction}} = 28.29 \times 2.00 \times \cos(180^\circ) = -56.58 \text{ J} \]

Step 4: Calculate Work Done by Gravity

The component of gravitational force acting along the incline is given by:\[ F_g = mg\sin\theta = 12.0 \times 9.8 \times \sin(53.0^\circ) \approx 94.08 \text{ N} \]Using the work formula \( W = f d \cos\phi \), substitute \( F_g \) for \( f \), and \( \phi = 0^\circ \):\[ W_{\text{gravity}} = 94.08 \times 2.00 \times \cos(0^\circ) = 188.16 \text{ J} \]

Step 5: Calculate Work Done by Normal Force

The normal force acts perpendicular to the motion, so the angle \( \phi = 90^\circ \).The work done by the normal force (\( W = f d \cos\phi \)) is:\[ W_{\text{normal}} = N \cdot d \cdot \cos(90^\circ) = 70.72 \times 2.00 \times 0 = 0 \text{ J} \]

Step 6: Calculate Net Work Done on the Package

The net work done is the sum of the works calculated: \( W_{\text{friction}} + W_{\text{gravity}} + W_{\text{normal}} \).\[ W_{\text{net}} = -56.58 + 188.16 + 0 = 131.58 \text{ J} \]

Key Concepts

Kinetic Friction

When objects like a package slide on surfaces, they often face resistance. This resistance is due to kinetic friction. Kinetic friction is a type of friction that acts when two surfaces move past each other. It always opposes the direction of motion.

The intensity of kinetic friction is determined by two main factors:

• The coefficient of kinetic friction (\( \mu_k \)), a number that represents how much friction two materials exhibit against each other.
• The normal force (\( N \)), which is the force perpendicular to the surfaces in contact.

To find the force of kinetic friction (\( f_k \)), we use the formula: \[ f_k = \mu_k \cdot N \]. Here, knowing that \( \mu_k = 0.40 \) and \( N = 70.72 \) N gives us \( f_k \approx 28.29 \) N. Kinetic friction does negative work on an object because it opposes its motion. Hence, when calculating work done, the angle between the force and the direction of motion is \( 180^\circ \). The calculated work done by kinetic friction is \( -56.58 \) J, indicating it lessens the energy available for motion.

Inclined Plane

An inclined plane is a flat surface tilted at an angle relative to the horizontal. It's one of the classic simple machines that makes moving objects up or down easier compared to lifting them vertically.

In our problem, the package slides down a chute tilted at \( 53.0^\circ \). This tilt alters the force of gravity, splitting it into two components:

• One acting perpendicular to the plane.
• Another acting parallel to the slope.

Gravity's component along the incline pulls the package downward. We calculate this force using \( F_g = mg\sin\theta \), where \( m = 12.0 \text{ kg} \), \( g = 9.8 \text{ m/s}^2 \), and \( \theta = 53.0^\circ \). This gives \( F_g \approx 94.08 \) N.
Tilted surfaces like inclined planes are crucial because they can change how forces act on objects. Here, gravity does \( 188.16 \) J of work, aiding the package's movement down the chute.

Net Work

Net work refers to the total work done on an object as it moves through a force field. It's the sum of all individual works from different forces acting upon the object.

In this exercise, three forces interact with the package:

• Friction, which does -56.58 J of work.
• Gravity, contributing 188.16 J of work.
• The normal force, which does no work, i.e., 0 J, as it acts perpendicular to the direction of motion.

To find the net work (\( W_{\text{net}} \)), sum these contributions: \[ W_{\text{net}} = -56.58 + 188.16 + 0 = 131.58 \text{ J} \].Net work tells us the change in the object's kinetic energy as it moves down the incline. This means the package gains energy due to gravity's dominating influence over the negative work of friction. Net work's significance lies in showing how different forces combine to affect the object’s overall state of motion.