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Chapter 7: Problem 51

A system of two paint buckets connected by a lightweight rope is released from rest with the \(12.0 \mathrm{~kg}\) bucket \(2.00 \mathrm{~m}\) above the floor (Fig. \(\mathbf{P 7 . 5 1}\) ). Use the principle of conservation of energy to find the speed with which this bucket strikes the floor. Ignore friction and the mass of the pulley.

Short Answer

Expert verified
The speed with which the bucket strikes the floor is approximately 7.85 m/s.

Step by step solution

01

Determine initial energy

Since the system is initially at rest, there is no kinetic energy. There is, however, potential energy driven by gravity. We calculate this as the product of mass (\(m=12kg\)), gravitational acceleration (\(g=9.8 m/s^2\)), and height (\(h=2m\)): So, the initial potential energy, \(PE_i = m*g*h = 12kg * 9.8 m/s^2 * 2m = 235.2 J\).
02

Use energy conservation principle

According to the principle of energy conservation, the total potential energy at the initial is equal to the total kinetic energy at the final. Because no other forces are acting on the buckets (friction and air resistance are ignored), their energy will only be converted from potential to kinetic. So the final kinetic energy \(KE_f = PE_i = 235.2 J\).
03

Calculate final speed

Now that we have the final kinetic energy, we can use the kinetic energy formula \( KE = 1/2 * m * v^2 \) to solve for the speed \(v\) (ignoring any loss due to friction or air resistance). Rearranging the formula for \(v\), we get \(v = \sqrt{(2* KE_f)/m} = \sqrt{(2*235.2 J)/12 kg}\).

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Key Concepts

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

Potential Energy
Potential energy is a crucial concept in physics that describes the energy stored in an object due to its position relative to a force. In the case of our exercise, we're dealing with gravitational potential energy. This form of potential energy is dependent on the object's mass, the height it is lifted to, and gravity's impact.
  • Mass (\(m\)): the amount of matter in the object. For the bucket, as stated, it's 12 kg.
  • Height (\(h\)): how far above the reference point (in this case, the floor) the object is. Here, \(h = 2 \text{m}\).
  • Gravitational acceleration (\(g\)): the rate at which an object will accelerate when falling. Always \(9.8 \text{m/s}^2\) on Earth's surface.
The formula for potential energy is represented as \(PE = m \cdot g \cdot h\). Substituting the given values, we find the potential energy of the bucket at its starting position to be 235.2 joules. This value captures all the energy stored in the bucket due to its elevated position.
Kinetic Energy
Kinetic energy refers to the energy of motion. When the bucket falls, its potential energy converts into kinetic energy. Kinetic energy is crucial in understanding how fast an object will move. Because energy is conserved throughout the process (ignoring air resistance and friction), the kinetic energy of the bucket as it approaches the floor matches the initial potential energy.
  • Mass (\(m\)): The same mass of the object (12 kg).
  • Velocity (\(v\)): This is the speed obtained due to the conversion of potential energy to kinetic energy.
Kinetic energy can be calculated using the formula \(KE = \frac{1}{2} \cdot m \cdot v^2\). Given \(KE = 235.2 \text{J}\), we rearrange the formula to find speed: \(v = \sqrt{\frac{2 \cdot KE}{m}}\). Solving this gives the bucket's velocity just before impact. This critical phase illustrates how stored energy transforms into the energy of motion.
Energy Conversion
Energy conversion addresses how energy transfers forms, maintaining overall conservation. In our paint bucket problem, this means potential energy transitioning into kinetic energy as the bucket falls. The principle of conservation of energy dictates that energy cannot simply disappear. Rather, it shifts from one form to another, explaining why the bucket's final speed links directly to its original height.
  • Conservation Principle: Total energy at the start equals total energy at the end.
  • No energy loss to air resistance or friction in this simplification.
By treating the system as ideal and neglecting friction, the only change is from potential to kinetic. Therefore, the energy conversion is a straightforward balancing act: \( PE_{initial} = KE_{final} \). This concept ensures that initial potential energy (height-based) fully becomes kinetic energy (speed-based). This constant conversion not only reinforces how energy behaves but also provides a practical demonstration of its application.

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Most popular questions from this chapter

A small block with mass \(m\) slides without friction on the inside of a vertical circular track that has radius \(R .\) What minimum speed must the block have at the bottom of its path if it is not to fall off the track at the top of its path?

The food calorie, equal to \(4186 \mathrm{~J},\) is a measure of how much energy is released when the body metabolizes food. A certain fruit-and-cereal bar contains 140 food calories. (a) If a 65 kg hiker eats one bar, how high a mountain must he climb to "work off" the calories, assuming that all the food energy goes into increasing gravitational potential energy? (b) If, as is typical, only \(20 \%\) of the food calories go into mechanical energy, what would be the answer to part (a)? (Note: In this and all other problems, we are assuming that \(100 \%\) of the food calories that are eaten are absorbed and used by the body. This is not true. A person's "metabolic efficiency" is the percentage of calories eaten that are actually used; the body eliminates the rest. Metabolic efficiency varies considerably from person to person.

A block with mass \(0.400 \mathrm{~kg}\) is on a horizontal frictionless surface and is attached to a horizontal compressed spring that has force constant \(k=200 \mathrm{~N} / \mathrm{m} .\) The other end of the spring is attached to a wall. The block is released, and it moves back and forth on the end of the spring. During this motion the block has speed \(3.00 \mathrm{~m} / \mathrm{s}\) when the spring is stretched \(0.160 \mathrm{~m}\). (a) During the motion of the block, what is its maximum speed? (b) During the block's motion, what is the maximum distance the spring is compressed from its equilibrium position? (c) When the spring has its maximum compression, what is the speed of the block and what is the magnitude of the acceleration of the block?

An ideal spring of negligible mass is \(12.00 \mathrm{~cm}\) long when nothing is attached to it. When you hang a \(3.15 \mathrm{~kg}\) weight from it, you measure its length to be \(13.40 \mathrm{~cm}\). If you wanted to store \(10.0 \mathrm{~J}\) of potential energy in this spring, what would be its total length? Assume that it continues to obey Hooke's law.

\(\mathrm{A} 1500 \mathrm{~kg}\) rocket is to be launched with an initial upward speed of \(50.0 \mathrm{~m} / \mathrm{s} .\) In order to assist its engines, the engineers will start it from rest on a ramp that rises \(53^{\circ}\) above the horizontal (Fig. \(\mathbf{P 7 . 5 0}\) ). At the bottom, the ramp turns upward and launches the rocket vertically. The engines provide a constant forward thrust of \(2000 \mathrm{~N}\), and friction with the ramp surface is a constant \(500 \mathrm{~N}\). How far from the base of the ramp should the \text { rocket start, as measured along the surface of the ramp?
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