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II An \(8.00 \mathrm{~kg}\) package in a mail-sorting room slides \(2.00 \mathrm{~m}\) 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?

Short Answer

Expert verified
(a) -37.8 J; (b) 125.4 J; (c) 0 J; (d) 87.6 J.

Step by step solution

01

Calculate the force of friction

The force of friction can be calculated using the formula:\[ f_k = \mu_k \times N \]Where \( f_k \) is the force of friction, \( \mu_k \) is the coefficient of kinetic friction (0.40), and \( N \) is the normal force. The normal force \( N \) can be found using:\[ N = m \cdot g \cdot \cos(\theta) \]Substitute the given values: \( m = 8.00 \mathrm{~kg} \), \( g = 9.81 \mathrm{~m/s^2} \), \( \theta = 53.0^\circ \).\[ N = 8.00 \times 9.81 \times \cos(53.0^\circ) \approx 47.2 \mathrm{~N} \]\[ f_k = 0.40 \times 47.2 \approx 18.9 \mathrm{~N} \]
02

Calculate the work done by friction

Work done by friction \( W_f \) is given by:\[ W_f = f_k \times d \times \cos(180^\circ) \]where \( d = 2.00 \mathrm{~m} \) is the distance slid, and \( \cos(180^\circ) = -1 \) because work done by friction is against the direction of motion. Therefore:\[ W_f = 18.9 \times 2.00 \times (-1) = -37.8 \mathrm{~J} \]
03

Calculate the work done by gravity

Work done by gravity \( W_g \) is calculated by:\[ W_g = m \cdot g \cdot d \cdot \sin(\theta) \]Substitute the values:\[ W_g = 8.00 \times 9.81 \times 2.00 \times \sin(53.0^\circ) \approx 125.4 \mathrm{~J} \]
04

Calculate the work done by the normal force

The work done by the normal force \( W_n \) on the package is zero because the normal force is perpendicular to the displacement. Thus:\[ W_n = 0 \mathrm{~J} \]
05

Calculate the net work done on the package

The net work done \( W_{net} \) is the sum of the work done by friction, gravity, and the normal force:\[ W_{net} = W_f + W_g + W_n \]\[ W_{net} = -37.8 + 125.4 + 0 = 87.6 \mathrm{~J} \]

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Key Concepts

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

Kinetic Friction
Kinetic friction is a type of friction that occurs when two surfaces slide past one another. It's important in physics because it can considerably affect the motion of objects. For an object on an inclined plane, kinetic friction acts to oppose its motion down the plane, which can alter the speed and energy of the object. It is calculated using the coefficient of kinetic friction (\( \mu_k \)) and the normal force (\( N \)). The force of kinetic friction (\( f_k \)) can be described using the formula \( f_k = \mu_k \times N \). In our problem, the force of kinetic friction is determined to be approximately 18.9 N, illustrating how the coefficient of friction and normal force combine to resist movement.
Inclined Plane Problems
Inclined plane problems are common exercises in physics that deal with objects moving along an inclined surface. Such problems involve breaking down forces acting along and perpendicular to the plane, allowing us to study motion in a more complex scenario than simple horizontal or vertical motion.
In our exercise, an 8.00 kg package slides down a 53.0° inclined plane. This setup involves calculating gravitational components and understanding how friction affects motion along the plane. Key aspects of solving inclined plane problems include:
  • Identifying the angle of inclination
  • Calculating normal force and its perpendicular nature
  • Determining horizontal and vertical components of forces like gravity
These elements help determine how an object moves and the energy changes involved.
Net Work Calculation
The concept of net work involves summing the total work done by all forces acting on an object. Net work can change the kinetic energy of an object and is crucial for predicting how its motion will evolve over time.
In this exercise, the work done by different forces (friction, gravity, and the normal force) is computed individually. The net work done on the package is calculated by adding all these contributions:\[W_{net} = W_f + W_g + W_n\]Net work is found to be 87.6 J, which signifies the energy transferred to or from the package due to all acting forces. This helps in understanding the overall effect of multiple interactions along the motion path.
Normal Force
The normal force is a support force that a surface exerts on an object resting upon it, and it acts perpendicular to the surface. In an inclined plane scenario, it is often less than the gravitational force acting on the object because it only counteracts the component of gravity perpendicular to the surface.
The normal force in this exercise is calculated using the mass of the package, gravitational acceleration, and the cosine of the inclination angle:\[ N = m \cdot g \cdot \cos(\theta) \]With these values, the normal force is approximately 47.2 N. Understanding the normal force is key, as it directly influences the kinetic friction acting on the object.
Work Done by Gravity
Work done by gravity is an essential part of understanding energy changes as an object moves along an inclined plane. Gravity acts downwards, and its component along the inclined plane helps in the movement of the object.
In this context, the work done by gravity is calculated with the product of mass, gravitational acceleration, distance moved, and the sine of the angle of inclination:\[W_g = m \cdot g \cdot d \cdot \sin(\theta) \]The calculated work done by gravity in the problem is approximately 125.4 J. This positive work indicates that gravity is doing work on the package, contributing to its energy as it slides down the chute.

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