Problem 6 - College Physics

Question

\(\cdot\) A 128.0 \(\mathrm{N}\) carton is pulled up a frictionless baggage ramp inclined at \(30.0^{\circ}\) above the horizontal by a rope exerting a 72.0 \(\mathrm{N}\) pull parallel to the ramp's surface. If the carton travels 5.20 \(\mathrm{m}\) along the surface of the ramp, calculate the work done on it by (a) the rope, (b) gravity, and (c) the normal force of the ramp. (d) What is the net work done on the carton? (e) Sup- pose that the rope is angled at \(50.0^{\circ}\) above the horizontal, instead of being parallel to the ramp's surface. How much work does the rope do on the carton in this case?

Step-by-Step Solution

Answer

(a) 374.4 J, (b) -332.8 J, (c) 0 J, (d) 41.6 J, (e) 351.3 J.

Step 1: Understanding Work Done

Work done on an object is given by the formula: \( W = F \cdot d \cdot \cos(\theta) \), where \( F \) is the force applied, \( d \) is the distance moved, and \( \theta \) is the angle between the force and direction of movement.

Step 2: Work Done by the Rope

For part (a), the force by the rope is parallel to the ramp, hence \( \theta = 0^{\circ} \). The work done by the rope is \( W = 72.0 \, \text{N} \times 5.20 \, \text{m} \times \cos(0^{\circ}) = 374.4 \, \text{J} \).

Step 3: Work Done by Gravity

For part (b), the gravitational force acts vertically, which is \(128.0 \, \text{N} \). The work done by gravity depends on the component of this force parallel to the ramp. The angle between the force of gravity and ramp is \(90^{\circ} - 30^{\circ} = 60^{\circ} \). Thus, \( W = 128.0 \, \text{N} \times 5.20 \, \text{m} \times \cos(60^{\circ}) = 332.8 \, \text{J} \), but since it's acting opposite to the motion, the actual work is \(-332.8 \, \text{J} \).

Step 4: Work Done by the Normal Force

For part (c), the normal force is perpendicular to the direction of motion, so \( \theta = 90^{\circ} \). Therefore, \( W = 128.0 \, \text{N} \times 5.20 \, \text{m} \times \cos(90^{\circ}) = 0 \, \text{J} \).

Step 5: Net Work Done on Carton

For part (d), the net work done is the sum of all individual work: \( 374.4 \, \text{J} - 332.8 \, \text{J} + 0 \, \text{J} = 41.6 \, \text{J} \).

Step 6: Work Done by Angled Rope

For part (e), if the rope is now pulling at \(50^{\circ}\) above the horizontal, the angle with the ramp is \(50^{\circ} - 30^{\circ} = 20^{\circ}\). Thus, the work done is \( W = 72.0 \, \text{N} \times 5.20 \, \text{m} \times \cos(20^{\circ}) \approx 351.3 \, \text{J} \).

Key Concepts

Inclined plane physics

Inclined planes are common examples of simple machines that help lift or lower objects with less effort. They consist of a flat surface tilted at an angle relative to the horizontal, which reduces the amount of force needed to do the same amount of work. By using inclined planes, we can explore how forces interact when objects move along these surfaces.
On an inclined plane, the force of gravity acts downwards, but the incline causes it to split into two components:

• Parallel to the surface of the incline (down the slope)
• Perpendicular to the surface of the incline (into the slope)

Whenever you deal with problems involving inclined planes, it's essential to understand how these force components affect the object in question. Understanding these interactions makes it easier to solve physics problems involving ramps and similar scenarios.

Work done calculation

Calculating work in physics involves three crucial factors: the force applied, the distance over which it is applied, and the angle between the force and the direction of movement. The work done is mathematically represented by the formula:

• \[ W = F \cdot d \cdot \cos(\theta) \]
• where:
  • \( W \) is the work done.
  • \( F \) is the force involved.
  • \( d \) is the distance the object moves.
  • \( \theta \) is the angle between the force and the direction of motion.

To break it down further, if the force is parallel to the direction of motion, \( \theta = 0^{\circ} \) and \( \cos(0^{\circ}) = 1 \), leading to maximum work. Conversely, when the force is perpendicular to the movement, as with the normal force, no work is done because \( \cos(90^{\circ}) = 0 \). Understanding how each force component contributes to work done is crucial in calculating it accurately, especially when dealing with inclined planes.

Forces on an incline

Forces acting on an inclined plane include gravitational force, normal force, and any applied forces, such as pulling or pushing activities. Gravity acts straight down, but an inclined surface splits it into two useful components:

• One component that acts parallel to the incline, causing the object to potentially slide downwards.
• The other component acts perpendicular to the surface, which the normal force balances out. The normal force is exerted by the surface to support the weight of the object perpendicular to the inclined plane.

The gravitational component parallel to the ramp can be calculated as:

• \( F_{\parallel} = F_{gravity} \cdot \sin(\text{angle}) \)

And the perpendicular component as:

• \( F_{\perp} = F_{gravity} \cdot \cos(\text{angle}) \)

When solving physics problems with inclined planes, identifying these force components is vital to understanding the system's overall mechanics.

Physics problem solving

Solving physics problems, especially those involving inclined planes, requires careful analysis and application of concepts and formulas. A step-by-step approach often makes tackling these problems more manageable:
First, identify all the forces acting on the object, such as gravity, normal force, and any horizontal or pulling forces. Next, determine the components of these forces along the inclined surface, which involves breaking them into parallel and perpendicular components. After that, apply the concept of work done using the formula:

• \[ W = F \cdot d \cdot \cos(\theta) \].

Calculate the work done by each force individually, considering the direction of motion relative to each force. Sum these values to find the net work done on the object.
Lastly, ensure that the correct angle is used in calculations. Sometimes, forces are applied at angles different from the incline, requiring recalculation of effective angles.
By following these structured steps, students can demystify complex problems related to work and energy on inclined planes, leading to a better understanding of physics concepts.