/*! 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 A \(0.800-\mathrm{kg}\) ball is ... [FREE SOLUTION] | 91Ó°ÊÓ

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A \(0.800-\mathrm{kg}\) 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 anywere, 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
(a) Both zero; (b) Tension zero, gravity 25.088 J.

Step by step solution

01

Analyze Part (a)(i) - Work Done by Tension in One Complete Circle

When the ball is swung in a vertical circle, the tension in the string is always perpendicular to the direction of displacement. Since work done is calculated using the formula \( W = F \cdot d \cdot \cos(\theta) \) and \( \theta = 90^\circ \) for tension, \( \cos(90^\circ) = 0 \). Thus, the total work done by the tension in one complete circle is zero.
02

Analyze Part (a)(ii) - Work Done by Gravity in One Complete Circle

Gravity does work on the ball as it moves in a circle. However, in one complete circle, the ball starts and ends at the same height, meaning the gravitational potential energy change is zero. Therefore, the work done by gravity over one complete circle is also zero since the potential energy change is zero.
03

Analyze Part (b)(i) - Work Done by Tension from Lowest to Highest Point

In a semicircle from the lowest to the highest point, the argument remains the same. Tension acts perpendicular to the displacement along the path at every point. Thus, the total work done by the tension in the string is also zero.
04

Calculate Part (b)(ii) - Work Done by Gravity from Lowest to Highest Point

Gravity does work on the ball as it moves from the lowest to highest point. The work done can be calculated using the change in gravitational potential energy: \[ W = mgh \]where \( m = 0.800 \text{ kg} \), \( g = 9.8 \text{ m/s}^2 \), \( h = 2 \times 1.60 \text{ m} = 3.2 \text{ m} \). Thus, \[ W = 0.800 \times 9.8 \times 3.2 = 25.088 \text{ J} \].

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Key Concepts

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

Tension in Physics
Tension is a force exerted through a string, rope, or cable when it is pulled tight by forces acting from opposite ends. In physics, when dealing with objects in motion, tension is an important concept, especially in circular motion problems. The key thing to remember is that the tension in a string or rope when an object moves in a vertical circle is always directed towards the center of the circle.
However, when calculating work done by tension, its direction matters greatly. Work done is defined by the equation \( W = F \cdot d \cdot \cos(\theta) \), where \( F \) is the force, \( d \) is the displacement, and \( \theta \) is the angle between the force and displacement directions.
In circular motion, especially in uniform circular motion, tension always acts perpendicular to displacement. Since \( \cos(90^\circ) = 0 \), the work done by tension over a complete circle or any arc of the circular path is zero. This key fact means:
  • No energy transfer occurs via tension when moving an object in a complete circle.
  • Tension does however provide the centripetal force crucial for maintaining the circular path.
Gravitational Potential Energy
Gravitational potential energy (GPE) is the energy an object possesses due to its position in a gravitational field. In simpler terms, it is the energy stored in an object when it is raised to a height h above a reference point. The formula for gravitational potential energy is given by \( U = mgh \), where \( m \) is mass, \( g \) is the acceleration due to gravity, and \( h \) is the height above the chosen reference point.
This concept becomes important when analyzing objects in vertical circular motion. When an object like a ball tied to a string moves in a vertical circle, it experiences changes in height relative to the ground, and hence changes in its gravitational potential energy.
Overall, understanding GPE helps in analyzing the work done by gravity. As the ball completes a full circle, its starting and ending height are the same, resulting in no net change in GPE, and thus no net work done by gravity. However, when the ball moves from the lowest point to the highest point in a semicircle, its height increases, and therefore the gravitational potential energy also increases.
  • Gravitational work done equals the change in gravitational potential energy.
  • No net gravitational work over a full circle due to zero net change in height.
Vertical Circular Motion
Vertical circular motion refers to the movement of an object along a circular path in a vertical plane. This type of motion combines principles of both linear and rotational dynamics, and requires an understanding of several forces acting on the object. In the instance of a ball on a string, both tension from the string and the gravitational force act on the ball.
While moving in a vertical circle, the velocity of the ball can vary significantly between the lowest and highest points, due to the work-energy relationship. At the highest point, the ball's speed is typically lower due to the higher potential energy, while at the lowest point, its speed is higher due to lower potential energy.
It's essential to comprehend that:
  • Centripetal force, provided by tension and gravity, is required to keep the object moving in a circle.
  • The force of tension varies at different points along the path, generally decreasing as the ball moves upwards and increasing as it moves downwards.
Understanding these forces aids in predicting the motion path and the object's speed at different points of the circle, providing a comprehensive perspective on the complex interplay of forces in vertical circular motion.

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Most popular questions from this chapter

A 5.00-kg package slides 1.50 \(\mathrm{m}\) down a long ramp that is inclined at \(12.0^{\circ}\) below the horizontal. The coefficient of kinetic friction between the package and the ramp is \(\mu_{\mathrm{k}}=0.310\) . Calculate (a) the work done on the package by friction; (b) the work done on the package by gravity, (c) the work done on the package by the normal force; (d) the total work done on the package. (e) If the package has a speed of 2.20 \(\mathrm{m} / \mathrm{s}\) at the top of the ramp, what is its speed after sliding 1.50 \(\mathrm{m}\) down the ramp?

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A physics student spends part of her day walking between classes or for recreation, during which time she expends energy at an average rate of 280 W. The remainder of the day she is sitting in class, studying, or resting; during these activities, she expends energy at an average rate of 100 W. If she expends a total of \(1.1 \times 10^{7} \mathrm{J}\) of energy in a 24 -bour day, how much of the day did she spend walking?

Proton Bombardment. A proton with mass \(1.67 \times10^{-27} \mathrm{kg}\) is propelled at an initial speed of \(3.00 \times 10^{5} \mathrm{m} / \mathrm{s}\) directiy 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 pro- ton get? (c) What is the speed of the proton when it is again 5.00 \(\mathrm{m}\) away from the uranium nucleus?

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