Problem 53 - University Physics with Modern Physics

Question

A 0.300-kg potato is tied to a string with length 2.50 m, and the other end of the string is tied to a rigid support. The potato is held straight out horizontally from the point of support, with the string pulled taut, and is then released. (a) What is the speed of the potato at the lowest point of its motion? (b) What is the tension in the string at this point?

Step-by-Step Solution

Answer

(a) 7.00 m/s; (b) 8.82 N

Step 1: Analyzing the System at the Lowest Point

At the lowest point of the potato’s motion, all of its potential energy has been converted to kinetic energy. We start by calculating the potential energy at the initial position. The height at the initial position is equal to the length of the string, which is 2.50 m.

Step 2: Calculating Initial Potential Energy

The initial gravitational potential energy (PE_initial) is given by:\[ PE_{initial} = m imes g imes h \]where, \( m = 0.300 \, \text{kg} \), \( g = 9.8 \, \text{m/s}^2 \), and \( h = 2.50 \, \text{m} \).\[ PE_{initial} = 0.300 imes 9.8 imes 2.50 = 7.35 \, \text{J} \].

Step 3: Using Conservation of Energy

By conservation of energy, the kinetic energy (KE) at the lowest point is equal to the initial potential energy:\[ KE = PE_{initial} = 7.35 \, \text{J} \].The formula for kinetic energy is:\[ KE = \frac{1}{2} m v^2 \].

Step 4: Solving for Speed

Rearranging the kinetic energy formula to solve for speed \(v\):\[ v = \sqrt{\frac{2 imes KE}{m}} \].Substitute \( KE = 7.35 \, \text{J} \) and \( m = 0.300 \, \text{kg} \):\[ v = \sqrt{\frac{2 imes 7.35}{0.300}} = \sqrt{49} = 7.00 \, \text{m/s} \].

Step 5: Calculating Tension in the String

At the lowest point, the forces acting on the potato are its weight and the tension in the string. The net force is centripetal, acting towards the center of the circle:\[ T - mg = \frac{mv^2}{r} \].Rearrange the equation for tension \(T\):\[ T = mg + \frac{mv^2}{r} \].

Step 6: Substituting Values into Tension Equation

Substitute \( m = 0.300 \, \text{kg} \), \( g = 9.8 \, \text{m/s}^2 \), \( v = 7.00 \, \text{m/s} \), and \( r = 2.50 \, \text{m} \) into the tension equation:\[ T = (0.300)(9.8) + \frac{0.300 \times 7.00^2}{2.50} \].Calculate:\[ T = 2.94 + \frac{0.300 \times 49}{2.50} \].\[ T = 2.94 + 5.88 = 8.82 \, \text{N} \].

Key Concepts

Conservation of Energy

The idea of conservation of energy is fundamental in physics, stating that energy cannot be created or destroyed. It can only change forms. In this scenario, a potato attached to a string is raised to a certain height. Initially, all its energy is in the form of gravitational potential energy. When the potato begins its descent, the potential energy converts into kinetic energy as it gains speed.

Understanding that the total energy (potential + kinetic) remains constant provides a powerful tool for solving problems. In this exercise, the potential energy at the highest point becomes kinetic energy at the lowest point. Therefore, by calculating the potential energy at the top, we can determine the kinetic energy (and hence the speed) at the bottom.

• Energy is conserved: potential energy converts to kinetic energy.
• No energy is lost in an ideal system.

Recognizing energy transformation helps simplify complicated problems in dynamics and motions under force.

Gravitational Potential Energy

Gravitational potential energy (GPE) is the energy an object possesses because of its position in a gravitational field. This energy depends on three factors:

• The mass of the object (m).
• The height above the reference point (h).
• The gravitational constant (g) which is approximately 9.8 m/s² on Earth.

The formula for GPE is \( PE = mgh \). In this exercise, the potato is initially 2.50 meters above its lowest point. We use the formula to calculate the initial potential energy, allowing us to determine how much energy will be converted into kinetic energy at the lowest point. By understanding how to calculate potential energy, you can predict how much energy is available for conversion to other forms, like kinetic energy.

Kinetic Energy

Kinetic energy is the energy of motion. When the potato swings down, its potential energy is converted into kinetic energy. The formula for kinetic energy is \( KE = \frac{1}{2}mv^2 \). Here, \( m \) is the mass, and \( v \) is the velocity of the object.

At the lowest point of motion, all potential energy has transformed into kinetic energy, giving us the speed we need to calculate. Rearranging the kinetic energy formula allows us to solve for velocity.\
The idea is simple but powerful: all the energy due to its height becomes energy due to its speed as it falls. By learning how to work with kinetic energy, you can solve problems that involve calculating speeds and understanding motion dynamics.

Centripetal Force

Centripetal force is required to keep an object moving in a circular path. This inward force is crucial when objects, like our swinging potato, move along a circular path. The force isn't a separate type of force but is provided by other forces (like tension in the string in this case) present in the system.

• Centripetal force is directed towards the center of the circular path.
• It is calculated using the formula \( F_{c} = \frac{mv^2}{r} \), where \( v \) is the velocity of the object, and \( r \) is the radius of the circle.

In this problem, tension in the string provides the centripetal force necessary to keep the potato moving in a circular arc. By understanding how tension combines with gravitational force (weight of the potato), you can calculate the total force acting at the lowest point in its path. This gives insights into how force dynamics arise in rotational motion and how centripetal forces manage circular motion efficiently.