Warning: foreach() argument must be of type array|object, bool given in /var/www/html/web/app/themes/studypress-core-theme/template-parts/header/mobile-offcanvas.php on line 20

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

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with Vaia!

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.

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

Two tugboats pull a disabled supertanker. Each tug exerts a constant force of 1.80 \(\times\) 10\(^6\) N, one 14\(^\circ\) west of north and the other 14\(^\circ\) east of north, as they pull the tanker 0.75 km toward the north. What is the total work they do on the supertanker?

A 6.0-kg box moving at 3.0 m/s on a horizontal, frictionless surface runs into a light spring of force constant 75 N/cm. Use the work\(-\)energy theorem to find the maximum compression of the spring.

A 5.00-kg block is moving at \(\upsilon_0\) \(=\) 6.00 m/s along a frictionless, horizontal surface toward a spring with force constant \(k\) = 500 N/m that is attached to a wall (\(\textbf{Fig. P6.79}\)). The spring has negligible mass. (a) Find the maximum distance the spring will be compressed. (b) If the spring is to compress by no more than 0.150 m, what should be the maximum value of \(\upsilon_0\)?

A proton with mass 1.67 \(\times\) 10\(^{-27}\) kg is propelled at an initial speed of 3.00 \(\times\) 10\(^5\) m/s directly toward a uranium nucleus 5.00 m away. The proton is repelled by the uranium nucleus with a force of magnitude \(F = \alpha/x^2\), where \(x\) is the separation between the two objects and \(\alpha = 2.12 \times 10^{-26} \, \mathrm{N} \cdot \mathrm{m}^2\). Assume that the uranium nucleus remains at rest. (a) What is the speed of the proton when it is \(8.00 \times 10^{-10}\) m from the uranium nucleus? (b) As the proton approaches the uranium nucleus, the repulsive force slows down the proton until it comes momentarily to rest, after which the proton moves away from the uranium nucleus. How close to the uranium nucleus does the proton get? (c) What is the speed of the proton when it is again 5.00 m away from the uranium nucleus?

A car is traveling on a level road with speed \(v_0\) at the instant when the brakes lock, so that the tires slide rather than roll. (a) Use the work energy theorem to calculate the minimum stopping distance of the car in terms of \(v_0\), \(g\), and the coefficient of kinetic friction \(\mu_k\) between the tires and the road. (b) By what factor would the minimum stopping distance change if (i) the coefficient of kinetic friction were doubled, or (ii) the initial speed were doubled, or (iii) both the coefficient of kinetic friction and the initial speed were doubled?

See all solutions

Recommended explanations on Physics Textbooks

View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.

Sign-up for free