/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 9 \(\mathrm{A} 0.800\) -kg ball is... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

\(\mathrm{A} 0.800\) -kg ball is tied to the end of a string 1.60 \(\mathrm{m}\) long and swung in a vertical circle. (a) During one complete circle, starting anywhere, calculate the total work done on the ball by (i) the tension in the string and (ii) gravity. (b) Repeat part (a) for motion along the semicircle from the lowest to the highest point on the path.

Short Answer

Expert verified
Complete circle: both zero. Semicircle: tension zero, gravity -25.1 J.

Step by step solution

01

Understanding the Problem

We need to determine the total work done on the ball by the tension in the string and by gravity for one complete circle and for a semicircle from the lowest to the highest point. Let's consider these separately for parts (a) and (b).
02

Work Done by Tension (Complete Circle)

In circular motion, the tension in the string does not perform work on the ball because the tension force is always perpendicular to the direction of displacement at every point in the circle. Therefore, for one complete circle, the work done by tension is:\[W_T = 0 \, \text{Joules}\]
03

Work Done by Gravity (Complete Circle)

The gravitational force is conservative, and the displacement is closed (starting and ending at the same position). The net work done by gravity over one complete circle is zero:\[W_G = 0 \, \text{Joules}\]
04

Work Done by Tension (Semicircle)

Similar to the full circle case, the work done by tension in moving the ball from the lowest to the highest point (semicircle) is still zero because the tension is always perpendicular to the displacement:\[W_T = 0 \, \text{Joules}\]
05

Work Done by Gravity (Semicircle)

When moving from the lowest point to the highest point, the work done by gravity can be calculated from the change in gravitational potential energy, as the force of gravity is conservative. The height difference is equal to the diameter of the circle (2 times the radius, r = 1.60 m), so:\[\Delta h = 2 \times 1.60 = 3.20 \, \text{m}\]The work done by gravity:\[W_G = -mgh = -(0.800 \, \text{kg})(9.81 \, \text{m/s}^2)(3.20 \, \text{m}) = -25.1 \, \text{Joules}\]
06

Conclusion

For the complete circle, work done by tension is zero and also zero by gravity. For the semicircle from the lowest to the highest point, work done by tension is zero, and by gravity it is -25.1 Joules.

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 91Ó°ÊÓ!

Key Concepts

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

Work Done by Tension
In a vertical circular motion, the tension in the string is a central force that constantly changes direction to keep the object moving in a circular path. While the tension is critical for maintaining the path, it does not contribute to any work being done on the object. The reason lies in the definition of work itself. Work done by a force is calculated using the formula: - Work = Force × Displacement × cos(θ) Here, θ is the angle between the force and the displacement. For tension, this angle is always 90 degrees since the tension in the string acts towards the center, while the displacement is along the circumference of the path. Since the cosine of 90 degrees is zero: - The work done by tension = Force × Displacement × cos(90) = 0 This holds true for all parts of the circle, whether it's a complete circle or just a semicircle. Thus, for the scenario presented: - The work done by tension is always zero.
Work Done by Gravity
Gravity plays a crucial role in circular motion, especially when the path is vertical. It is considered a conservative force, which means the work done by gravity is related only to the change in height, irrespective of the path taken. For a complete circle: - The object returns to its initial height. Thus, the net change in height is zero. Consequently, the work done by gravity across a full circle is zero because there is no overall displacement in the vertical direction. In the specific case of a semicircle: - The object moves from the lowest point to the highest point. - This involves a definitive change in height, which equals twice the radius of the circular path (since the lowest to the highest point is the diameter). The work done by gravity is given by the expression: - Work done by gravity = -mgh Where: - **m** is the mass of the object, - **g** is the acceleration due to gravity (9.81 m/s²), and - **h** is the change in height, which is the diameter here (2 × 1.60 m = 3.20 m). Thus, when moving from the lowest to the highest point: - The work done by gravity = -0.800 kg × 9.81 m/s² × 3.20 m = -25.1 Joules. This negative work signifies that gravity acts in the direction opposite to the displacement, opposing the motion.
Gravitational Potential Energy
Gravitational potential energy (GPE) is energy stored due to an object's position in a gravitational field. In circular motion, especially when dealing with vertical circles, changes in height affect an object's GPE. The formula for gravitational potential energy is: - GPE = mgh Where: - **m** is the mass of the object, - **g** is the gravitational acceleration (9.81 m/s²), and - **h** is the height relative to a reference point. In a vertical circular path: - At the lowest point, the GPE is at its minimum because the height is the least relative to the highest point. - At the highest point, the height is maximum, hence the GPE is at its maximum. The change in gravitational potential energy as an object moves from the lowest to the highest point reflects the work done against gravity. This energy change is what results in the negative work calculated earlier. It's important to understand that, during this motion: - Energy is conserved but transforms from kinetic to potential and vice versa. Gaining a clear grasp of gravitational potential energy is vital as it connects the concepts of force, work, and energy in motion.

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

A student proposes a design for an automobile crash barrier in which a 1700 -kg sport utility vehicle moving at 20.0 \(\mathrm{m} / \mathrm{s}\) crashes into a spring of negligible mass that slows it to a stop. So that the passengers are not injured, the acceleration of the vehicle as it slows can be no greater than 5.00\(g .\) (a) Find the required spring constant \(k,\) and find the distance the spring will compress in slowing the vehicle to a stop. In your calculation, disregard any deformation or crumpling of the vehicle and the friction between the vehicle and the ground. (b) What disadvantages are there to this design?

BIO Chin-Ups, While doing a chin-up, a man lifts his body 0.40 \(\mathrm{m} .\) (a) How much work must the man do per kilogram of body mass? (b) The muscles involved in doing a chin-up can generate about 70 \(\mathrm{J}\) of work per kilogram of muscle mass. If the man can just barely do a \(0.40-\mathrm{m}\) chin-up, what percentage of his body's mass do these muscles constitute? (For comparison, the total percentage of muscle in a typical \(70-\mathrm{kg}\) man with 14\(\%\) body fat is about 43\(\%\) . (c) Repeat part (b) for the man's young son, who has arms half as long as his father's but whose muscles can also generate 70 \(\mathrm{J}\) of work per kilogram of muscle mass. (d) Adults and children have about the same percentage of muscle in their bodies. Explain why children can commonly do chin-ups more easily than their fathers.

Automotive Power I. A truck engine transmits 28.0 \(\mathrm{kW}(37.5 \mathrm{hp})\) to the driving wheels when the truck is traveling at a constant velocity of magnitude 60.0 \(\mathrm{km} / \mathrm{h}\) (37.3 \(\mathrm{mi} / \mathrm{h} )\) on a level road. (a) What is the resisting force acting on the truck? (b) Assume that 65\(\%\) of the resisting force is due to rolling friction and the remainder is due to air resistance. If the force of rolling friction is independent of speed, and the force of air resistance is proportional to the square of the speed, what power will drive the truck at 30.0 \(\mathrm{km} / \mathrm{h} ?\) At 120.0 \(\mathrm{km} / \mathrm{h} ?\) Give your answers in kilowatts and in horsepower.

CALC Proton Bombardment. A proton with mass \(1.67 \times 10^{-27} \mathrm{kg}\) is propelled at an initial speed of \(3.00 \times 10^{5} \mathrm{m} / \mathrm{s}\) directly toward a uranium nucleus 5.00 \(\mathrm{m}\) 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 \(8.00 \times 10^{-10} \mathrm{m}\) 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 \(\mathrm{m}\) away from the uranium nucleus?

Some Typical Kinetic Energies. (a) In the Bohr model of the atom, the ground- state electron in hydrogen has an orbital speed of 2190 \(\mathrm{km} / \mathrm{s} .\) What is its kinetic energy? (Consult Appendix F.) (b) If you drop a 1.0-kg weight (about 2 lb) from a height of 1.0 \(\mathrm{m}\) , how many joules of kinetic energy will it have when it reaches the ground? (c) Is it reasonable that a \(30-\mathrm{kg}\) child could run fast enough to have 100 \(\mathrm{J}\) of kinetic energy?

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.