Problem 11 - University Physics with Modern Physics

Question

A 128.0-N carton is pulled up a frictionless baggage ramp inclined at 30.0\(^\circ\) above the horizontal by a rope exerting a 72.0-N pull parallel to the ramp's surface. If the carton travels 5.20 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) Suppose 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) 352.5 J

Step 1: Work Done by the Rope Parallel to the Ramp

To calculate the work done by the rope, use the formula: \[ W_{rope} = F_{rope} \times d \times \cos(\theta) \]where \( F_{rope} = 72.0 \text{ N} \), \( d = 5.20 \text{ m} \), and \( \theta = 0\) degrees since the force is parallel to the ramp. Substituting these values, we have:\[ W_{rope} = 72.0 \times 5.20 \times \cos(0) = 374.4 \text{ J} \].

Step 2: Work Done by Gravity

The work done by gravity is calculated using:\[ W_{gravity} = m \cdot g \cdot h \]where \( m \cdot g = 128.0 \text{ N} \) is the weight of the carton, and \( h = d \cdot \sin(30.0^\circ) \) is the height change.Calculating the height change:\[ h = 5.20 \times \sin(30.0^\circ) = 2.60 \text{ m} \].Then, work done by gravity is:\[ W_{gravity} = 128.0 \times -2.60 = -332.8 \text{ J} \].(We take it as negative since gravity opposes the motion up the incline.)

Step 3: Work Done by the Normal Force

The normal force acts perpendicular to the motion along the ramp. Since work is only done by a force component parallel to displacement, the work done by the normal force is zero:\[ W_{normal} = 0 \text{ J} \].

Step 4: Net Work Done on the Carton

The net work done on the carton is the sum of the work done by all forces:\[ W_{net} = W_{rope} + W_{gravity} + W_{normal} \].Substitute the values:\[ W_{net} = 374.4 - 332.8 + 0 = 41.6 \text{ J} \].

Step 5: Work Done by the Rope at 50 degrees Angle

When the rope is at 50 degrees above horizontal, the tension must be resolved into components parallel and perpendicular to the ramp.The parallel component is:\[ F_{rope, parallel} = 72.0 \times \cos(20.0^\circ) \] (where 20.0° is the angle between the force and the ramp).Calculate the work done using the parallel component:\[ W_{rope, angled} = F_{rope, parallel} \times 5.20 = 72.0 \times \cos(20.0^\circ) \times 5.20 \approx 352.5 \text{ J} \].

Key Concepts

Inclined Plane Mechanics

Inclined plane mechanics deal with how objects interact with flat surfaces that are tilted relative to the horizontal. These surfaces or planes simplify many real-world scenarios, like ramps or slides, into manageable physics problems. In our exercise, you are asked to find the work done on a carton being pulled on a frictionless inclined plane. The angle of inclination is crucial as it affects how forces act on the object. A key angle here is 30.0°, which determines how much of the gravitational force opposes the motion of the carton. The inclined plane acts as a medium to resolve forces acting along two main directions:

• Parallel to the incline (along the plane)
• Perpendicular to the incline (against the plane)

Understanding the mechanics of the inclined plane helps predict how much effort is needed to move an object up or down the ramp against different forces acting upon it.

Force Components

In physics, breaking forces into components can simplify understanding of how an object moves. The key to solving inclined plane problems is identifying which component of a force works parallel to the plane. For example, when the rope pulls the carton upwards, the force exerted by the rope can initially be split into components that are parallel and perpendicular to the plane. For a force parallel to the slope, like the carton's upward pull, we use:

• The force itself completely works in line with movement along the slope.
• The cosine of the angle between force direction and movement becomes one.

When the rope's angle changes, the effective force gets calculated from its components:

• The parallel component becomes the actual contributing factor for motion along the plane.
• The new angle introduces an adjustment, using \( \theta = 50^\circ - 30^\circ \, to achieve the effective angle for solving the work done by the rope.
• We calculate using the formula: \ (F_{rope, parallel} = F \times \cos(\text{adjusted angle}) \)

Understanding the vector components is essential for comprehensively interpreting the effects of each force in terms of physics principles.

Net Work Calculation

Net work calculation involves summing up all individual works done by forces acting on an object. In simpler terms, net work is the total energy change resulting from various forces acting upon an object along a specific path. Understanding this helps us characterize how efficiently energy is being used in object movement or task performance. For the carton moving along the ramp:

• Work by rope utilizes rope's entire parallel force, yielding positive work.
• Gravity's work component is negative, opposing the carton's upward travel.
• The normal force acts perpendicular and thus contributes zero to net work, as its force component doesn't cause displacement.

The central concept of a net work makes us appreciate how energies from differing sources accumulate or cancel out. It helps determine the overall motion's energy aspect, allowing calculations of efficiencies, speeds, and projections.