Problem 63 - University Physics with Modern Physics

Question

A luggage handler pulls a 20.0 -kg suitcase up a ramp inclined at \(25.0^{\circ}\) above the horizontal by a force \(\vec{\boldsymbol{F}}\) 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 \(\vec{\boldsymbol{F}} ;\) (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; (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 \(\mathrm{m}\) along the ramp?

Step-by-Step Solution

Answer

(a) 532 J, (b) -315 J, (c) 0 J, (d) -201.44 J, (e) 15.56 J, (f) 1.25 m/s.

Step 1: Work done by the force \( \vec{\boldsymbol{F}} \)

The work done by a force is given by the formula \( W = Fd\cos(\theta) \), where \( F \) is the force, \( d \) is the distance traveled, and \( \theta \) is the angle between the force and the direction of motion. Since the force acts parallel to the ramp, \( \theta = 0 \). Substitute \( F = 140 \text{ N} \), \( d = 3.80 \text{ m} \), and \( \cos(0) = 1 \), so:\[ W = 140 \times 3.80 \times 1 = 532 \text{ J} \]

Step 2: Work done by the gravitational force

The gravitational force does work that is perpendicular to the ramp. The component of the gravitational force along the ramp is \( mg \sin \theta \), where \( m = 20.0 \text{ kg} \), \( g = 9.8 \text{ m/s}^2 \), and \( \theta = 25.0^{\circ} \). The work done by gravity is:\[ W = -(mg \sin \theta) \times d = -(20 \times 9.8 \times \sin 25^{\circ}) \times 3.80 \]Calculate \( \sin 25^{\circ} \) and substitute:\[ W = -(196 \times 0.4226) \times 3.80 \approx -315 \text{ J} \]

Step 3: Work done by the normal force

The normal force acts perpendicular to the direction of motion, so the work done by the normal force is zero. Hence:\[ W = 0 \text{ J} \]

Step 4: Work done by the friction force

The friction force is \( f_k = \mu_k N \), where the normal force \( N = mg \cos \theta \). With \( \mu_k = 0.300 \), compute \( N \) first:\[ N = mg \cos \theta = 20 \times 9.8 \times \cos 25^{\circ} \]Calculate \( \cos 25^{\circ} \) and substitute:\[ N = 20 \times 9.8 \times 0.9063 = 176.7 \text{ N} \]Thus, \( f_k = 0.300 \times 176.7 = 53.01 \text{ N} \). The friction opposes the motion, thus the work is:\[ W = -(f_k) \times d = -(53.01) \times 3.80 = -201.44 \text{ J} \]

Step 5: Total work done on the suitcase

The total work \( W_{\text{tot}} \) is the sum of the works of all contributing forces:\[ W_{\text{tot}} = 532 - 315 + 0 - 201.44 = 15.56 \text{ J} \]

Step 6: Calculate final speed using the work-energy principle

The work-energy principle states that the change in kinetic energy \( \Delta KE \) is equal to the total work done \( W_{\text{tot}} \). If the suitcase starts from rest, its initial kinetic energy \( KE_i = 0 \). Thus,\[ \Delta KE = \frac{1}{2}mv^2 - 0 = W_{\text{tot}} \]Solve for the final velocity \( v \):\[ \frac{1}{2} \times 20 \times v^2 = 15.56 \]\[ 10v^2 = 15.56 \]\[ v^2 = 1.556 \]\[ v = \sqrt{1.556} \approx 1.25 \text{ m/s} \]

Key Concepts

Kinetic Friction

Kinetic friction opposes the motion of an object sliding over a surface. It's a force that resists movement, making it harder for objects to slide smoothly. In our scenario, a 20 kg suitcase is being pulled up a ramp, and the force of kinetic friction is acting against its motion.

• The coefficient of kinetic friction (\( \mu_k\) ) between the suitcase and the ramp is given as 0.300. This value is crucial as it determines how much frictional force will act on the suitcase.
• The frictional force (\( f_k\) ) is calculated using the formula \[ f_k = \mu_k \cdot N \], where \( N\) is the normal force. In this context, the frictional force is calculated to be approximately 53.01 N.
• The work done by the kinetic friction force is negative because it acts in the direction opposite to the displacement of the suitcase. This turns into a value of \(-201.44\) Joules, indicating that energy is "lost" in overcoming friction.

Awareness of kinetic friction helps in understanding energy loss due to resistance in various physics problems.
Understanding this force is key to solving problems involving motion on surfaces, as it affects the energy and acceleration of moving objects.

Gravitational Force

Gravitational force is the force of attraction between two masses, in this case, the Earth and the suitcase. It always acts downwards, towards the center of the Earth.

• The gravitational force acting on an object is given by \( mg \), where \( m \) is the mass of the object and \( g \) is the acceleration due to gravity, approximately \( 9.8 \, \text{m/s}^2 \).
• However, when considering an inclined plane, the component of this gravitational force along the ramp is \( mg \sin(\theta) \). For the suitcase moving up an inclined ramp at \( 25^{\circ} \), this force works opposite to its movement.
• This component is responsible for the work done by the gravitational force, which is calculated as \(-315 \) Joules. The negative sign indicates that gravity does work against the movement of the suitcase.

Understanding gravitational force helps illustrate how weight influences movement, especially on inclined surfaces.
Gravitational pull is crucial in calculating net forces and understanding potential energy dynamics.

Normal Force

The normal force is a contact force that acts perpendicular to the surface with which an object is in contact. It provides balance against the gravitational pull pushing the object into the surface.

• For objects on flat surfaces, the normal force \( N \) is equal to the gravitational force \( mg \). However, on an incline, it is slightly less, calculated as \( mg \cos(\theta) \).
• In our case for a \( 25^{\circ} \) incline, \( N \) comes out to be about 176.7 N. This value is essential for calculating forces like friction that depend on the normal force.
• The key takeaway here is that the work done by a force acting perpendicular to the direction of motion, such as the normal force, is zero. That's why the normal force doesn't contribute to the net work done on the suitcase.

Understanding normal force is essential because it directly impacts the magnitude of kinetic friction and ensures accurate force balance calculations in motion problems involving inclined planes.
It helps in solving equilibrium problems and influences the effective weight management of objects on surfaces.