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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

A block with mass \(m=\) \(\begin{array}{lll}0.200 \mathrm{~kg} & \text { is placed against a com- }\end{array}\) pressed spring at the bottom of a ramp that is at an angle of \(53.0^{\circ}\) above the horizontal. The spring has \(8.00 \mathrm{~J}\) of elastic potential energy stored in it. The spring is released, and the block moves up the incline. After the block has traveled a distance of \(3.00 \mathrm{~m},\) its speed is \(4.00 \mathrm{~m} / \mathrm{s} .\) What is the magnitude of the friction force that the ramp exerts on the block while the block is moving?

Two blocks are attached to either end of a light rope that passes over a light, frictionless pulley suspended from the ceiling. One block has mass \(8.00 \mathrm{~kg},\) and the other has mass \(6.00 \mathrm{~kg}\). The blocks are released from rest. (a) For a \(0.200 \mathrm{~m}\) downward displacement of the \(8.00 \mathrm{~kg}\) block, what is the change in the gravitational potential energy associated with each block? (b) If the tension in the rope is \(T\), how much work is done on each block by the rope? (c) Apply conservation of energy to the system that includes both blocks. During the \(0.200 \mathrm{~m}\) downward displacement, what is the total work done on the system by the tension in the rope? What is the change in gravitational potcntial energy associated with the system? Use energy conservation to find the speed of the \(8.00 \mathrm{~kg}\) block after it has descended \(0.200 \mathrm{~m} .\)

\(\mathrm{A} 0.150 \mathrm{~kg}\) block of ice is placed against a horizontal, compressed spring mounted on a horizontal tabletop that is \(1.20 \mathrm{~m}\) above the floor. The spring has force constant \(1900 \mathrm{~N} / \mathrm{m}\) and is initially compressed \(0.045 \mathrm{~m}\). The mass of the spring is negligible. The spring is released, and the block slides along the table, goes off the edge, and travels to the floor. If there is negligible friction between the block of ice and the tabletop, what is the speed of the block of ice when it reaches the floor?

During the calibration process, the cantilever is observed to deflect by \(0.10 \mathrm{nm}\) when a force of \(3.0 \mathrm{pN}\) is applied to it. What deflection of the cantilever would correspond to a force of \(6.0 \mathrm{pN} ?\) (a) \(0.07 \mathrm{nm}\) (b) \(0.14 \mathrm{nm} ;\) (c) \(0.20 \mathrm{nm} ;\) (d) \(0.40 \mathrm{nm}\).

A 1.20 kg piece of cheese is placed on a vertical spring of negligible mass and force constant \(k=1800 \mathrm{~N} / \mathrm{m}\) that is compressed \(15.0 \mathrm{~cm} .\) When the spring is released, how high does the cheese rise from this initial position? (The cheese and the spring are not attached.)
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