Problem 10 - University Physics with Modern Physics

Question

An 8.00 -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?

Step-by-Step Solution

Answer

(a) \(-50.3\ J\), (b) \(78.7\ J\), (c) \(0\ J\). Net work = \(28.4\ J\).

Step 1: Calculate the Force of Friction

The force of friction can be calculated using the formula \( F_{friction} = \mu_k \times F_{normal} \). First, solve for the normal force \( F_{normal} \). Since the package is on an incline, the normal force is given by \( F_{normal} = mg \cos \theta \), where \( m = 8.00 \ kg \) is the mass, \( g = 9.81 \ m/s^2 \) is the acceleration due to gravity, and \( \theta = 53.0^{\circ} \) is the angle of the incline. So, \( F_{normal} = 8.00 \times 9.81 \times \cos(53.0^{\circ}) \). Then calculate \( F_{friction} = 0.40 \times F_{normal} \).

Step 2: Calculate the Work Done by Friction

The work done by friction is given by the formula \( W_{friction} = F_{friction} \times d \times \cos(180^{\circ}) \). Since friction opposes the direction of motion, the angle between the force of friction and the displacement is \(180^{\circ}\), making the cosine negative. Substitute \( d = 2.00 \ m \) to find \( W_{friction} \).

Step 3: Calculate the Work Done by Gravity

The work done by gravity can be calculated using \( W_{gravity} = F_{gravity} \times d \times \cos(\theta) \). Here, \( F_{gravity} = mg \sin \theta \) which represents the component of gravitational force along the incline. Substitute \( m = 8.00 \ kg \), \( g = 9.81 \ m/s^2 \), \( \theta = 53.0^{\circ} \), and \( d = 2.00 \ m \) to find \( W_{gravity} = mg \sin \theta \times d \).

Step 4: Calculate the Work Done by the Normal Force

The work done by the normal force is given by \( W_{normal} = F_{normal} \times d \times \cos(90^{\circ}) \). Since the normal force is perpendicular to the displacement, the cosine of \(90^{\circ}\) is zero, so \( W_{normal} = 0 \).

Step 5: Calculate the Net Work Done on the Package

The net work done on the package is the sum of the work done by friction, gravity, and the normal force: \( W_{net} = W_{friction} + W_{gravity} + W_{normal} \). Use the values calculated in the previous steps to compute the net work.

Key Concepts

Kinetic Friction

Kinetic friction occurs when two surfaces slide against one another while moving. This friction is always in the opposite direction of the object's motion, acting to slow it down. In our exercise, we have an 8.00 kg package moving down a chute with kinetic friction. The force of friction is proportional to the normal force and the coefficient of kinetic friction (which is 0.40 here). The formula for kinetic friction is:

• \( F_{friction} = \mu_k \times F_{normal} \)

The coefficient of kinetic friction, \( \mu_k \), is a dimensionless number reflecting the friction level between the surfaces. A higher \( \mu_k \) means more friction, requiring more force to keep the object moving or make it move faster.
In this scenario, kinetic friction acts to reduce the package's kinetic energy as it slides down. Despite this resistive force, the package continues its journey down the plane, illustrating how friction influences objects in motion.

Inclined Plane

An inclined plane is a simple surface tilted at an angle to the horizontal. This angle affects how forces act on an object. Our problem deals with a 53.0° incline. The inclined plane affects both the normal force and gravitational components working on the package.

• Normal Force is reduced on an incline.
• Gravitational force has components along and perpendicular to the plane.

The angle directly influences these components. When an object is on an incline, gravity can be broken down into two components: one parallel to the slope, causing the object to slide, and one perpendicular, affecting the normal force. This breakup of forces leads to different calculations and behaviors compared to when the object is on a flat surface.

Normal Force

Normal force is the perpendicular force exerted by a surface to support an object resting on it. On flat surfaces, it equals the gravitational force. However, things change on an inclined plane. Here, the normal force is less because it's tilted away from the full gravitational force acting perpendicular to the surface.For our package sliding down the incline:

• The normal force is given by: \( F_{normal} = mg \cos \theta \)

This formula shows how the angle \( \theta \) (53.0° in this case) affects the normal force. As the angle increases, the portion of gravity opposing the normal force decreases. It's essential in calculating kinetic friction since friction's force depends on the normal force. Despite the incline, the work done by the normal force is zero in this scenario because it acts perpendicular to the motion of the package.

Work Done by Gravity

Work done by gravity is a calculation of how much energy gravity transfers as an object moves. This concept is crucial in understanding motion on an inclined plane. The formula for work done by gravity is:

• \( W_{gravity} = F_{gravity} \times d \times \cos(\theta) \)

In our example, the force of gravity along the incline is \( F_{gravity} = mg \sin \theta \). Gravity acts parallel to the plane to slide the package downwards. So, \( \theta \) in this context is the incline angle, showing how much of gravity's force works along the path the object travels.
The component of gravitational force along the plane gets multiplied by the distance to calculate the total work done by gravity. This work contributes positively to the object's movement, contrasting with kinetic friction, which resists it. Understanding how gravity performs work helps in appreciating how energy changes form and function as objects move along an inclined plane.