Question

A luggage handler pulls a 20.0-kg suitcase up a ramp inclined at 32.0\(^\circ\) above the horizontal by a force \(\overrightarrow{F}\) of magnitude 160 N that acts parallel to the ramp. The coefficient of kinetic friction between the ramp and the incline is \(\mu_k = 0.300\). If the suitcase travels 3.80 m along the ramp, calculate (a) the work done on the suitcase by \(\overrightarrow{F}\); (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; (e) 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 m along the ramp?

Short answer

(a) 608 J, (b) -393.3 J, (c) 0 J, (d) -189.6 J, (e) 25.1 J, (f) 1.58 m/s.

Step-by-step solution

Determine the work done by the force F

The work done by a force acting parallel to the displacement is given by the formula \( W = F \cdot d \cdot \cos(\theta) \), where \( \theta \) is the angle between the force and the direction of displacement. Since the force \( F = 160 \text{ N} \) acts parallel to the ramp, \( \theta = 0 \) and \( \cos(0) = 1 \). The displacement \( d = 3.80 \text{ m} \). So, the work done by \( \overrightarrow{F} \) is:\[W_F = 160 \text{ N} \times 3.80 \text{ m} \times 1 = 608 \text{ J}\]

Calculate the work done by the gravitational force

The work done by gravity can be calculated using the component of the gravitational force along the ramp. The gravitational force acting on the suitcase is \( mg \), and the component along the ramp is \( mg \sin(\theta) \). Thus, the work done by gravity \( W_g \) is:\[W_g = mg \sin(\theta) \cdot d = 20.0 \text{ kg} \times 9.8 \text{ m/s}^2 \times \sin(32^\circ) \times 3.80 \text{ m}\]\[= 20.0 \times 9.8 \times 0.5299 \times 3.80 = 393.3 \text{ J}\]This work is negative because gravity acts in the opposite direction to the displacement.

Work done by the normal force

The normal force acts perpendicular to the direction of the displacement. Therefore, the work done by the normal force is zero, as work is only done by forces with components parallel to the displacement. Thus:\[ W_n = 0 \text{ J} \]

Determine the work done by friction

Frictional force opposes the motion, and its magnitude is given by \( f_k = \mu_k N \), where \( N = mg \cos(\theta) \) is the normal force. Hence:\[f_k = 0.300 \times 20.0 \times 9.8 \times \cos(32^\circ) = 49.9 \text{ N}\]The work done by friction is then given by:\[W_f = -f_k \cdot d = -49.9 \times 3.80 = -189.6 \text{ J}\]The negative sign indicates the friction opposes the motion.

Calculate the total work done on the suitcase

The total work done on the suitcase is the sum of all the individual work components:\[W_{\text{total}} = W_F + W_g + W_n + W_f = 608 - 393.3 + 0 - 189.6 = 25.1 \text{ J}\]

Determine the final speed of the suitcase

Using the work-energy principle, \[W_{\text{total}} = \Delta KE = \frac{1}{2}mv^2 - \frac{1}{2}m(0)^2\]Thus,\[25.1 = \frac{1}{2} \times 20.0 \times v^2\]Solving for \( v \):\[v^2 = \frac{2 \times 25.1}{20.0}\]\[v = \sqrt{2.51} = 1.58 \text{ m/s}\]

Key concepts

Kinetic Friction

Kinetic friction is a type of friction that occurs between moving surfaces. When an object slides over a surface, this force resists the motion, working opposite to the direction of movement. In our scenario, the suitcase experiences kinetic friction as it slides up the ramp.
To calculate the force of kinetic friction, we use the formula:

• Frictional Force (\(f_k\)) = Coefficient of Kinetic Friction (\(\mu_k\)) \(\times\) Normal Force (\(N\)).

For the inclined plane, the normal force goes perpendicular to the surface. We express the normal force as \(N = mg \cdot \cos(\theta)\), with \(m\) being the mass, \(g\) the acceleration due to gravity, and \(\theta\) the angle of the incline.
The kinetic frictional force, therefore, becomes: \(f_k = \mu_k \cdot mg \cdot \cos(\theta)\).
It's important to note that the work done by kinetic friction is negative because it opposes the direction of the suitcase's motion, dissipating energy as heat.

Gravitational Force

Gravitational force is the attractive force that pulls the suitcase toward the Earth's center. In the case of an inclined plane, this force can be divided into two components: one parallel and one perpendicular to the plane's surface.
The component of gravitational force working parallel to the incline is crucial for calculating work done against gravity. This component is given by:

• Parallel Component: \(mg \cdot \sin(\theta)\)

Where \(m\) is the suitcase's mass, \(g\) represents gravitational acceleration, and \(\theta\) is the incline's angle.
In terms of the work-energy principle, the work done by the gravitational force as the suitcase moves up the incline is calculated by multiplying this force component by the distance moved, with a negative sign specifying that it opposes the direction of motion:

• Work done by gravity: \(W_g = -mg \cdot \sin(\theta) \cdot d\)

Here, \(d\) is the distance traveled along the slope, and the negative sign indicates that gravitational force opposes the suitcases' movement up the ramp.

Inclined Plane

An inclined plane is a flat surface tilted at an angle relative to the horizontal's flat surface. It is a simple machine used to ease the effort needed to raise or lower objects. When dealing with physics problems involving inclined planes, we typically focus on decomposing forces along the plane and perpendicular to it.
The angle of the incline, \(\theta\), plays a crucial role in determining how different forces act on an object. For instance, the steeper the angle, the greater the component of gravitational force pulling an object down the slope. In contrast, the normal force, which acts perpendicularly to the surface, decreases with steeper angles.
To solve problems involving inclined planes:

• Break down forces into components along and perpendicular to the incline.
• Use trigonometric functions like sine and cosine for calculations.
• Consider forces like friction, gravity, and any applied force in your equations.

Understanding inclined planes is vital for analyzing scenarios where forces and motions occur on slopes, ensuring that all force components are accurately calculated and considered.

Normal Force

The normal force is a supportive force exerted by a surface perpendicular to the object resting on it. On an inclined plane, unlike horizontal surfaces, the normal force doesn't equal the object's weight. Instead, it is lesser because only the component of weight perpendicular to the incline influences it.
To compute the normal force on an inclined plane, we use the relationship: \(N = mg \cdot \cos(\theta)\).
This equation indicates:

• \(N\) is the normal force.
• \(m\) refers to the mass of the object.
• \(g\) represents the gravitational acceleration.
• \(\theta\) is the incline's angle.

The normal force is critical in calculating kinetic friction, as frictional force depends on it. Unlike forces contributing to work done, the normal force itself does no work in this scenario as it is perpendicular to the direction of motion.