91Ó°ÊÓ

A luggage handler pulls a \(20.0-\mathrm{kg}\) suitcase up a ramp inclined at \(25.0^{\circ}\) above the horizontal by a force \(\overrightarrow{\boldsymbol{F}}\) of magnitude 140 \(\mathrm{N}\) that acts parallel to the ramp. The coefficient of kinetic friction between the ramp and the incline is \(\mu_{\mathrm{k}}=0.300\) . If the suitcase travels 3.80 \(\mathrm{m}\) along the ramp, calculate (a) the work done on the suitcase by the force \(\overrightarrow{\boldsymbol{F}},\) (b) the work done on the suitcase by the gravitational force; (c) the work done on the suitcase by the normal force; (d) the work done on the suitcase by the friction force; (c) the total work done on the suitcase. (f) If the speed of the suitcase is zero at the bottom of the ramp, what is its speed after it has traveled 3.80 \(\mathrm{m}\) m along the ramp?

Step by step solution

01

Calculate Work Done by Applied Force

The work done by a force is given by the formula: \[ W = F \cdot d \cdot \cos(\theta) \]where \( F \) is the magnitude of the force, \( d \) is the distance, and \( \theta \) is the angle between the force and direction of motion. Since the force is parallel to the incline (\( \theta = 0 \)), we have:\[ W_F = 140 \times 3.80 \times \cos(0^{\circ}) = 140 \times 3.80 = 532 \, \text{J} \]

02

Calculate Work Done by Gravitational Force

The gravitational force component doing work is \( mg \sin(\alpha) \), where \( \alpha = 25^{\circ} \) and \( m = 20.0 \, \text{kg} \). The work done by gravity is:\[ W_g = m \cdot g \cdot d \cdot \sin(\alpha) = 20.0 \times 9.81 \times 3.80 \times \sin(25^{\circ}) \]Calculating gives:\[ W_g = -314 \, \text{J} \]

03

Calculate Work Done by Normal Force

The normal force acts perpendicular to the displacement. Hence, the work done by the normal force is zero because the angle between the normal force and the displacement is \( 90^{\circ} \):\[ W_n = 0 \]

04

Calculate Work Done by Frictional Force

The force of friction is given by \( f_k = \mu_k \cdot N \), where \( N = mg \cos(\alpha) \). Substitute and calculate this:\[ f_k = 0.300 \times 20.0 \times 9.81 \times \cos(25^{\circ}) \]Calculating the frictional force:\[ f_k \approx 53.1 \, \text{N} \]The work done by friction, acting opposite to the distance, is:\[ W_f = -f_k \cdot d = -53.1 \times 3.80 = -202 \, \text{J} \]

05

Calculate Total Work Done

The total work done on the suitcase is the sum of the work done by all the forces:\[ W_{total} = W_F + W_g + W_n + W_f = 532 - 314 + 0 - 202 \approx 16 \, \text{J} \]

06

Calculate Final Speed Using Work-Energy Principle

Using the work-energy principle, the total work done is equal to the change in kinetic energy:\[ \Delta KE = \frac{1}{2} mv^2 - \frac{1}{2} m(0)^2 = W_{total} \]\[ \frac{1}{2} \times 20.0 \times v^2 = 16 \]Solve for \( v \):\[ v^2 = \frac{16 \times 2}{20.0} = 1.6 \]\[ v = \sqrt{1.6} \approx 1.26 \, \text{m/s} \]

07

Complete Exercise

All parts of the exercise are now solved: work by the applied force (532 J), work by gravity (-314 J), work by the normal force (0 J), work by friction (-202 J), total work (16 J), and final speed (1.26 m/s).

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Key Concepts

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

Kinetic Friction

When an object moves across a surface, it encounters a force called kinetic friction. Kinetic friction acts parallel to the surface and opposes the direction of motion. It is determined by

• the coefficient of kinetic friction (\( \mu_k \)), which depends on the materials involved, and
• the normal force (N), which is the perpendicular force exerted by the surface on the object.

The formula for kinetic friction is \( f_k = \mu_k \cdot N \). In situations where an incline is involved, the normal force is calculated by considering the component of gravitational force perpendicular to the incline, \( N = mg \cos(\alpha) \), where \( \alpha \) is the incline angle.

A key fact about kinetic friction is that it does negative work because it acts opposite to motion. This means kinetic friction removes energy from the system. In this context, when the suitcase is pulled up the ramp, the kinetic friction works against the applied pulling force, which makes it crucial to account for it when calculating the total work done.

Work Done by Forces

Work is a measure of energy transfer when a force moves an object over a distance. The basic formula for work done by a constant force is \( W = F \cdot d \cdot \cos(\theta) \), where:

• \( F \) is the force applied,
• \( d \) is the distance over which the force is applied, and
• \( \theta \) is the angle between the force direction and the direction of movement.

Different forces can perform work differently depending on how they are applied:

• **Applied Force**: If the force is applied parallel to the direction of movement, \( \theta \) is zero, and work is maximized (\( \cos(0^{\circ}) = 1 \)).
• **Gravitational Force**: When climbing an incline, the gravitational force has a component parallel to the incline, which does negative work since it's opposite to the uphill motion.
• **Normal Force**: Acts perpendicular to the movement; thus, it does no work as \( \theta = 90^{\circ} \) and \( \cos(90^{\circ}) = 0 \).

By understanding how each force contributes to the total work done, students can better grasp how energy is conserved or dissipated in mechanical systems.

Inclined Plane Mechanics

Inclined planes are surfaces tilted at an angle to the horizontal. They are common in physics problems and help to learn about forces and motion. When analyzing motion on an inclined plane:

• The angle of inclination \( \alpha \) affects the components of gravitational force. The component acting along the incline is \( mg \sin(\alpha) \), and the perpendicular component is \( mg \cos(\alpha) \).
• Kinetic friction must be considered as well, factored in by the frictional force formula based on the normal force.

Inclined plane problems often include calculating potential and kinetic energy changes. The work-energy principle is central, stating that the total work done is equal to the change in kinetic energy:\[ \Delta KE = \frac{1}{2} mv^2 - \frac{1}{2} m(v_0)^2\]
In many situations, calculating the total sum of work done by forces acting on an object will tell you how its kinetic energy changes, letting you find its final speed. Understanding these fundamental mechanics equips students with problem-solving tools for various real-world engineering and physics applications.