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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?

Short Answer

Expert verified
(a) 374.4 J, (b) -332.8 J, (c) 0 J, (d) 41.6 J, (e) 352.5 J

Step by step solution

01

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:\[ W_{rope} = 72.0 \times 5.20 \times \cos(0) = 374.4 \text{ J} \].
02

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.)
03

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:\[ W_{normal} = 0 \text{ J} \].
04

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:\[ W_{net} = 374.4 - 332.8 + 0 = 41.6 \text{ J} \].
05

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:\[ W_{rope, angled} = F_{rope, parallel} \times 5.20 = 72.0 \times \cos(20.0^\circ) \times 5.20 \approx 352.5 \text{ J} \].

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Key Concepts

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

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.

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Most popular questions from this chapter

A mass \(m\) slides down a smooth inclined plane from an initial vertical height \(h\), making an angle \(\alpha\) with the horizontal. (a) The work done by a force is the sum of the work done by the components of the force. Consider the components of gravity parallel and perpendicular to the surface of the plane. Calculate the work done on the mass by each of the components, and use these results to show that the work done by gravity is exactly the same as if the mass had fallen straight down through the air from a height \(h\). (b) Use the work\(-\)energy theorem to prove that the speed of the mass at the bottom of the incline is the same as if the mass had been dropped from height \(h\), independent of the angle \(\alpha\) of the incline. Explain how this speed can be independent of the slope angle. (c) Use the results of part (b) to find the speed of a rock that slides down an icy frictionless hill, starting from rest 15.0 m above the bottom.

All birds, independent of their size, must maintain a power output of 10\(-\)25 watts per kilogram of body mass in order to fly by flapping their wings. (a) The Andean giant hummingbird (\(Patagona gigas\)) has mass 70 g and flaps its wings 10 times per second while hovering. Estimate the amount of work done by such a hummingbird in each wingbeat. (b) A 70-kg athlete can maintain a power output of 1.4 kW for no more than a few seconds; the \(steady\) power output of a typical athlete is only 500 W or so. Is it possible for a human-powered aircraft to fly for extended periods by flapping its wings? Explain.

The human heart is a powerful and extremely reliable pump. Each day it takes in and discharges about 7500 L of blood. Assume that the work done by the heart is equal to the work required to lift this amount of blood a height equal to that of the average American woman (1.63 m). The density (mass per unit volume) of blood is \(1.05 \times 10^3 \, \mathrm{kg/m}^3\). (a) How much work does the heart do in a day? (b) What is the heart's power output in watts?

A physics professor is pushed up a ramp inclined upward at 30.0\(^\circ\) above the horizontal as she sits in her desk chair, which slides on frictionless rollers. The combined mass of the professor and chair is 85.0 kg. She is pushed 2.50 m along the incline by a group of students who together exert a constant horizontal force of 600 N. The professor's speed at the bottom of the ramp is 2.00 m/s. Use the work\(-\)energy theorem to find her speed at the top of the ramp.

We usually ignore the kinetic energy of the moving coils of a spring, but let's try to get a reasonable approximation to this. Consider a spring of mass \(M\), equilibrium length \(L_0\), and force constant \(k\). The work done to stretch or compress the spring by a distance \(L\) is \\(\frac{1}{2}\\) \(kX^2\), where \(X = L - L_0\). Consider a spring, as described above, that has one end fixed and the other end moving with speed \(v\). Assume that the speed of points along the length of the spring varies linearly with distance \(l\) from the fixed end. Assume also that the mass \(M\) of the spring is distributed uniformly along the length of the spring. (a) Calculate the kinetic energy of the spring in terms of \(M\) and \(v\). (\(Hint\): Divide the spring into pieces of length \(dl\); find the speed of each piece in terms of \(l\), \(v\), and \(L\); find the mass of each piece in terms of \(dl\), \(M\), and \(L\); and integrate from \(0\) to \(L\). The result is \(not\) \\(\frac{1}{2}\\) \(Mv^2\), since not all of the spring moves with the same speed.) In a spring gun, a spring of mass 0.243 kg and force constant 3200 N/m is compressed 2.50 cm from its unstretched length. When the trigger is pulled, the spring pushes horizontally on a 0.053-kg ball. The work done by friction is negligible. Calculate the ball's speed when the spring reaches its uncompressed length (b) ignoring the mass of the spring and (c) including, using the results of part (a), the mass of the spring. (d) In part (c), what is the final kinetic energy of the ball and of the spring?

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