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Chapter 7: Problem 55

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. 7.36\() .\) 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 final speed of the bucket is approximately 6.26 m/s.

Step by step solution

01

Understand the Problem

We have a system consisting of two buckets connected by a rope, with one of the buckets weighing 12.0 kg and initially positioned 2.00 meters above the ground. The goal is to find the speed of this bucket when it hits the ground, assuming no energy losses due to friction or the pulley鈥檚 mass.
02

Outline the Conservation of Energy Principle

The principle of conservation of energy states that the total energy in an isolated system remains constant. Here, the potential energy of the elevated bucket will be completely converted into the kinetic energy of the buckets when the 12.0 kg bucket reaches the floor.
03

Write the Energy Conservation Equation

Initially, the system has gravitational potential energy, given by \( U_i = mgh \), where \( m = 12.0 \text{ kg} \) and \( h = 2.00 \text{ m} \). The bucket's kinetic energy when it hits the floor is given by \( K_f = \frac{1}{2}mv_f^2 \). Thus, the energy conservation equation is: \[ mgh = \frac{1}{2}mv_f^2 \]
04

Solve for Final Speed

Solve the conservation equation for the final speed \( v_f \). Since the mass \( m \) appears on both sides of the equation, it cancels out. This gives us:\[ gh = \frac{1}{2}v_f^2 \]Rearranging and solving for \( v_f \), we get:\[ v_f = \sqrt{2gh} = \sqrt{2 \times 9.81 \text{ m/s}^2 \times 2.00 \text{ m}} \]
05

Calculate the Final Speed

Calculate the final speed using the values for \( g \) and \( h \):\[ v_f = \sqrt{2 \times 9.81 \times 2.00} = \sqrt{39.24} \approx 6.26 \text{ m/s} \]

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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 an important concept in physics that refers to the energy stored in an object due to its height above a reference point. In our context, it is tied to the 12 kg paint bucket suspended 2 meters above the floor. This type of energy is given by the formula: \( U = mgh \), where \( m \) is the mass of the object, \( g \) is the acceleration due to gravity (approximated to be 9.81 m/s虏 on Earth), and \( h \) is the height above the reference point.

In the given exercise: - The mass \( m \) is 12.0 kg. - The height \( h \) is 2.00 m. - Thus, the gravitational potential energy is calculated as \( U = 12.0 \times 9.81 \times 2.00 \).

This potential energy represents how much work the gravitational force can do on the bucket as it falls, transforming that energy into other forms, such as kinetic energy. Understanding gravitational potential energy helps in solving many real-world physics problems involving heights and brought into play whenever objects are lifted to a certain height.
Kinetic Energy
Kinetic energy is the energy an object possesses due to its motion. When the paint bucket, initially at rest, begins to fall, its potential energy is converted into kinetic energy.

Kinetic energy is expressed by the formula: \( K = \frac{1}{2}mv^2 \), where \( m \) is the mass of the object and \( v \) is its velocity.

For the falling bucket: - The mass \( m \) remains 12.0 kg. - As it falls, it gains speed, and the kinetic energy increases until it reaches the floor. - The formula implies that even a small increase in velocity results in a significant increase in kinetic energy, due to the squaring of \( v \).

In the exercise, conservation of energy principle allows us to equate the initial potential energy to the final kinetic energy, leading to calculating the speed of the bucket when it strikes the floor. This speed is crucial for determining the kinetic energy at that moment.
Energy Conversion
Energy conversion is a process of changing one form of energy into another, and it plays a vital role in understanding how systems behave under the principle of conservation of energy.

In the case of the paint buckets, initially, the system possesses gravitational potential energy due to the elevated position of one bucket. As the bucket begins to fall, this stored potential energy is converted into kinetic energy.

Here is how the conversion process unfolds: - At the start, with the bucket elevated, the system's energy is entirely potential. - As the bucket falls, the potential energy decreases while the kinetic energy increases. - If there were no other forces (like friction) acting on the system, the potential energy would be completely converted into kinetic energy by the time the bucket hits the ground.

The principle of energy conversion exemplifies how energy remains within a system but changes its form to accomplish motion or other effects. Understanding this conversion aids in predicting the behavior of dynamic systems and is foundational in physics education and practical applications.

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Most popular questions from this chapter

You are asked to design a spring that will give a \(1160-\mathrm{kg}\) satellite a speed of 2.50 \(\mathrm{m} / \mathrm{s}\) relative to an orbiting space shuttle. Your spring is to give the satellite a maximum acceleration of 5.00 \(\mathrm{g}\) . The spring's mass, the recoil kinetic energy of the shuttle, and changes in gravitational potential energy will be negligible. (a) What must the force constant of the spring be? (b) What distance must the spring be compressed?

Gravity in Two Dimensions. Two point masses, \(m_{1}\) and \(m_{2},\) lie in the \(x y\) -plane, with \(m_{1}\) held in place at the origin and \(m_{2}\) free to move a distance \(r\) away at a point \(P\) having coordinates \(x\) and \(y\) (Fig. 7.27\()\) . The gravitational potential energy of these masses is found to be \(U(r)=-G m_{1} m_{2} / r,\) where \(G\) is the gravitational constant. (a) Show that the components of the force on \(m_{2}\) due to \(m_{1}\) are $$ F_{x}=-\frac{G m_{1} m_{2} x}{\left(x^{2}+y^{2}\right)^{3 / 2}} \quad \text { and } \quad F_{y}=-\frac{G m_{1} m_{2} y}{\left(x^{2}+y^{2}\right)^{3 / 2}} $$ (Hint: First write \(r\) in terms of \(x\) and \(y . )(\text { b) Show that the magnitude}\) of the force on \(m_{2}\) is \(F=G m_{1} m_{2} / r^{2} .\) (c) Does \(m_{1}\) attract or repel \(m_{2} ?\) How do you know?

A force parallel to the \(x\) -axis acts on a particle moving along the \(x\) -axis. This force produces potential energy \(U(x)\) given by \(U(x)=\alpha x^{4},\) where \(\alpha=1.20 \mathrm{J} / \mathrm{m}^{4} .\) What is the force (magnitude and direction) when the particle is at \(x=-0.800 \mathrm{m} ?\)

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 \(550-\mathrm{N}\) kinetic friction force on each 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 8.0 \(\mathrm{m}\) along the ramp. Once stopped, a crate must not rebound back up the ramp. Calculate the force constant of the spring that will be needed in order to meet the design criteria.

A 72.0 -kg swimmer jumps into the old swimming hole from a diving board 3.25 \(\mathrm{m}\) above the water. Use energy conservation to find his speed just he hits the water (a) if he just holds his nose and drops in, (b) if he bravely jumps straight up (but just beyond the board) at 2.50 \(\mathrm{m} / \mathrm{s}\) , and (c) if he manages to jump downward at 2.50 \(\mathrm{m} / \mathrm{s} .\)
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