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Chapter 7: Problem 53

\(\mathrm{CP}\) A \(0.300 \mathrm{~kg}\) potato is tied to a string with length \(2.50 \mathrm{~m}\) 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?

Short Answer

Expert verified
The speed of the potato at the lowest point of its motion is 7.39 m/s. The tension in the string at that point is 3.31 N.

Step by step solution

01

Calculate Initial Potential Energy

The initial potential energy of the potato is given by the equation: \(PE = m * g * h\) where \(m = 0.300 kg\) is the mass of the potato, \(g = 9.81 m/s^2\) is the acceleration due to gravity, and \(h = 2.50 m\) is the height. Substituting these values into the equation gives \(PE = 0.300 kg * 9.81 m/s^2 * 2.50 m = 7.35 J\)
02

Calculate Final Kinetic Energy and Speed

At the lowest point of motion, all potential energy has been converted to kinetic energy. The kinetic energy is given by the equation: \(KE = 1/2 * m * v^2\). Setting \(KE\) equal to \(PE\) and solving for \(v\), we get: \(1/2 * 0.300 kg * v^2 = 7.35 J\). Solving for \(v\) gives \(v = \sqrt{(2 * 7.35 J) / 0.300 kg} = 7.39 m/s\)
03

Calculate Tension in the String

The tension in the string can be calculated by adding the centripetal force to the weight of the potato. The centripetal force is given by \(m * v^2 / r\). The weight of the potato is given by \(m * g\). Adding these quantities gives the tension \(T\): \(T = m * v^2 / r + m * g = 0.300 kg * (7.39 m/s)^2 / 2.50 m + 0.300 kg * 9.81 m/s^2 = 3.31 N\)

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Key Concepts

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

Potential Energy
When you hold a potato tied to a string horizontally, it stores energy due to its position. This stored energy is called potential energy. In the context of our problem, the potato's potential energy (PE) is determined by its mass, the height it's held at, and the gravitational force. The formula is given by:
  • \(PE = m \cdot g \cdot h\)
where \(m\) is the mass (0.300 kg), \(g\) is gravity (9.81 m/s虏), and \(h\) is the height of 2.50 m. So, holding the potato like this gives it 7.35 Joules of potential energy. This energy is what will transform into kinetic energy once the potato is let go.
Kinetic Energy
As the potato swings down, the potential energy changes into kinetic energy, which is the energy of motion. This conversion is crucial because it allows us to understand how fast something moves at the lowest point of its swing. Kinetic energy (KE) is given by the equation:
  • \(KE = \frac{1}{2} m v^2\)
At the lowest point, all potential energy has turned into kinetic energy. Therefore, we can set \(KE\) equal to the initial potential energy (7.35 J) and solve for the speed \(v\) of the potato. Doing the math gives us a speed of 7.39 m/s.
Centripetal Force
Centripetal force is essential when something is moving in a circular path. This force points towards the center of the circle, keeping the object moving in its path. In our scenario, the potato swings in a circular arc, meaning centripetal force comes into play. This force is calculated by:
  • \(F_c = \frac{m v^2}{r}\)
where \(m\) is the mass, \(v\) is the speed, and \(r\) is the string's length (2.50 m). At the lowest point, this centripetal force helps determine the total tension in the string.
Tension in Physics
Tension is the force experienced by the string in our pendulum setup as it supports the swinging potato. At the lowest point of the motion, the tension is not just the potato's weight but also includes the centripetal force because the potato is moving. The formula to find the tension \(T\) in the string is:
  • \(T = \frac{m v^2}{r} + m g\)
This equation sums the centripetal force and the gravitational force acting on the potato. As we calculate, this turns out to be 3.31 Newtons, showing how the motion and gravity combine to affect the tension at that point.

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Most popular questions from this chapter

You are designing a delivery ramp for crates containing exercise equipment. The \(1470 \mathrm{~N}\) crates will move at \(1.8 \mathrm{~m} / \mathrm{s}\) at the top of a ramp that slopes downward at \(22.0^{\circ} .\) The ramp exerts a \(515 \mathrm{~N}\) kinctic friction force on cach crate, and the maximum static friction force also has this value. Each crate will compress a spring at the bottom of the ramp and will come to rest after traveling a total distance of \(5.0 \mathrm{~m}\) along the ramp. Once stopped, a crate must not rebound back up the ramp. Calculate the largest force constant of the spring that will be needed to meet the design criteria.

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An ideal spring stores potential energy \(U_{0}\) when it is compressed a distance \(x_{0}\) from its uncompressed length. (a) In terms of \(U_{0}\). how much energy does the spring store when it is compressed (i) twice as much and (ii) half as much? (b) In terms of \(x_{0}\). how much must the spring be compressed from its uncompressed length to store (i) twice as much energy and (ii) half as much energy?

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