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A system of two paint buckets connected by a lightweight rope is released from rest with the 12.0 -kg bucket 2.00 \(\mathrm{m}\) above the floor (Fig. \(\mathrm{P} 7.55 ) .\) Use the principle of conservation of energy to find the speed with which this bucket strikes the floor. You can ignore friction and the mass of the pulley.

Short Answer

Expert verified
The bucket strikes the floor with a speed of 6.26 m/s.

Step by step solution

01

Identify the Energy Types and Initial Conditions

In this scenario, we have gravitational potential energy and kinetic energy involved. Initially, because the system is at rest, all energy is gravitational potential energy (due to the 12.0 kg bucket). Since the 12.0 kg bucket is 2.00 meters above the floor, gravitational potential energy is given by \( U = mgh \), where \( m = 12.0 \) kg, \( g = 9.8 \) m/s\(^2\), and \( h = 2.00 \) m.
02

Calculate Initial Potential Energy

Using the formula for gravitational potential energy \( U = mgh \), calculate:\[U = 12.0 \, \text{kg} \times 9.8 \, \text{m/s}^2 \times 2.00 \, \text{m} = 235.2 \, \text{J}\]
03

Set up the Conservation of Energy Equation

The principle of conservation of energy states that the total mechanical energy remains constant (assuming no friction). Therefore, initial energy (potential) equals final energy (kinetic) as the 12.0 kg bucket hits the ground:\[mgh = \frac{1}{2}mv^2\]where \( v \) is the speed of the bucket when it strikes the floor. Since the mass \( m \) appears on both sides, we can cancel it out.
04

Solve for the Final Speed

Substitute gravitational potential energy \( mgh = 235.2 \) J into the equation:\[235.2 = \frac{1}{2} \times 12.0 \times v^2\]Solving for \( v^2 \) gives:\[v^2 = \frac{235.2}{6.0} = 39.2\]Taking the square root of both sides, the speed \( v \) is:\[v = \sqrt{39.2} = 6.26 \, \text{m/s}\]
05

Conclusion

The 12.0 kg bucket will hit the floor with a speed of approximately 6.26 m/s according to the conservation of energy principles.

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Key Concepts

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

Gravitational Potential Energy
Gravitational potential energy is the energy an object possesses due to its position in a gravitational field. Imagine holding a paint bucket at a height above the ground; it has stored energy because of its elevated position. This energy is called gravitational potential energy (GPE), and it can be calculated using the formula \( U = mgh \), where:
  • \( m \) is the mass of the object (in kg)
  • \( g \) is the acceleration due to gravity (approximately \( 9.8 \, \text{m/s}^2 \) on Earth)
  • \( h \) is the height above the reference point (in meters)
In our exercise, a 12.0 kg bucket is initially 2.00 meters above the floor, so the potential energy is 235.2 Joules. The energy is potential because it is ready to be converted into another form, like kinetic energy, when the bucket is released.
Kinetic Energy
Kinetic energy is the energy an object has due to its motion. When the paint bucket falls, the gravitational potential energy it initially had is transformed into kinetic energy. This transformation is what makes the bucket accelerate towards the ground. The formula for kinetic energy (KE) is \( KE = \frac{1}{2}mv^2 \), where:
  • \( m \) is the mass of the object (in kg)
  • \( v \) is the velocity of the object (in meters per second)
As the bucket gains speed, its kinetic energy increases, reaching its maximum just before hitting the ground. Once you solve for the velocity, you see how potential energy becomes kinetic energy through motion.
Mechanical Energy
Mechanical energy is the sum of kinetic and potential energy in a system. It represents the total energy available to do work. In our scenario, the total mechanical energy at the start is purely gravitational potential energy.
Once the bucket begins to fall, this energy begins converting into kinetic energy while maintaining the same total mechanical energy level. Since no external forces like friction are acting in this simplified system, the conservation of mechanical energy principle applies. This means that the total energy before the fall equals the total energy just before the bucket hits the ground.
In simpler terms:
  • Start: Mechanical energy = Gravitational potential energy
  • End: Mechanical energy = Kinetic energy
This constant total energy illustrates key principles of physics.
Energy Transformation
Energy transformation is the process through which energy changes from one form to another. In the context of this exercise, energy transformation is observed from the moment the bucket is released until it hits the ground. Initially, all the system's energy is in the form of gravitational potential energy.
As the bucket begins to fall, this energy is gradually converted into kinetic energy, causing the bucket to accelerate. The energy transformation is complete when the bucket reaches the ground. At that point, the potential energy is zero, and all initial potential energy has transformed into kinetic energy, thereby increasing the speed of the bucket.
Understanding energy transformation helps in recognizing how energy is conserved and transmuted across different states, enabling you to predict and calculate outcomes in dynamic systems.

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Most popular questions from this chapter

Up and Down the Hill. \(\wedge 28\) -kg rock approaches the foot of a hill with a speed of 15 \(\mathrm{m} / \mathrm{s}\) . This hill slopes upward at a constant angle of \(40.0^{\circ}\) above the horizontal. The coefficients of static and kinetic friction between the hill and the rock are 0.75 and 0.20 , respectively. (a) Use energy conservation to find the maximum height above the foot of the hill reached by the rock. (b) Will the rock remain at rest at its highest point, or will it slide back down the hill? (c) If the rock does slide back down, find its speed when it returns to the bottom of the hill.

EP A 10.0 -kg microwave oven is pushed 8.00 m up the sloping surface of a loading ramp inclined at an angle of \(36.9^{\circ}\) above the horizontal, by a constant force \(\vec{\boldsymbol{F}}\) with a magnitude 110 \(\mathrm{N}\) and acting parallel to the ramp. The coefficient of kinetic friction between the oven and the ramp is 0.250 . (a) What is the work done on the oven by the force \(F ?\) (b) What is the work done on the oven by the friction force? (c) Compute the increase in potential energy for the oven. (d) Use your answers to parts (a), (b), and (c) to calculate the increase in the oven's kinetic energy. (e) Use \(\Sigma \vec{F}=m \vec{a}\) to calculate the acceleration of the oven. Assuming that the oven is initially at rest, use the acceleration to calculate the oven's speed after traveling 8.00 \(\mathrm{m}\) . From this, compute the increase in the oven's kinetic energy, and compare it to the answer you got in part (d).

Bi0 Food Calories. The food calorie, equal to 4186 \(\mathrm{J}\) . is a measure of how much energy is released when food is metabolized by the body. A certain brand of fruit-and-cereal bar contains 140 food calories per bar. (a) If a \(65-\mathrm{kg}\) hiker eats one of these bars, how high a mountain must he climb to "work off" the calories, assuming that all the food energy goes only 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 actually not true. A person's "metabolic efficiency" is the percentage of calories eaten that are actually used; the rest are eliminated by the body. Metabolic efficiency varies considerably from person to person.)

You are testing a new amusement park roller coaster with an empty car of mass 120 \(\mathrm{kg}\) . One part of the track is a vertical loop with radius 12.0 \(\mathrm{m}\) . At the bottom of the loop (point \(A )\) the car has speed \(25.0 \mathrm{m} / \mathrm{s},\) and at the top of the loop (point \(B )\) it has speed 8.0 \(\mathrm{m} / \mathrm{s} .\) As the car rolls from point \(A\) to point \(B\) , how much work is done by friction?

CALC (a) Is the force \(\vec{F}=C y^{2} \hat{\jmath},\) where \(C\) is a negative constant with units of \(\mathrm{N} / \mathrm{m}^{2}\) , conservative or nonconservative? Justify your answer. (b) Is the force \(\vec{\boldsymbol{F}}=C y^{2 \hat{\boldsymbol{l}}}\) , where \(C\) is a negative constant with units of \(\mathrm{N} / \mathrm{m}^{2},\) conservative or nonconservative? Justify your answer.

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