Intro to Conservation of Energy: Videos & Practice Problems

The conservation of mechanical energy states that the total mechanical energy (kinetic energy + potential energy) in an isolated system remains constant when only conservative forces act. For example, when a ball falls, its gravitational potential energy converts to kinetic energy without loss, maintaining a total energy of 1,960 joules. The principle can be expressed as Ki + Ui = Kf + Uf. Understanding this principle is crucial for solving various physics problems efficiently.

Conservation Of Mechanical Energy

The conservation of mechanical energy is a fundamental principle in physics that states the total mechanical energy of a system remains constant if only conservative forces are acting on it. This principle is particularly useful when analyzing the energy transformations of an object as it moves between different points, such as a ball falling from a height. The mechanical energy of a system is defined as the sum of its kinetic energy (K) and potential energy (U), expressed mathematically as:

$$E_{mechanical} = K + U$$

When considering a ball dropped from a height of 100 meters, we can calculate its mechanical energy at two points: the top (initial) and the bottom (final). At the top, the ball has gravitational potential energy due to its height, while its kinetic energy is zero since it starts from rest. The potential energy can be calculated using the formula:

$$U = mgh$$

where:

• m = mass of the object (in kilograms)
• g = acceleration due to gravity (approximately 9.8 m/s²)
• h = height above the reference point (in meters)

For the ball at 100 meters, the potential energy is:

$$U_{initial} = mg(100)$$

Assuming a mass of 2 kg, the potential energy would be:

$$U_{initial} = 2 \times 9.8 \times 100 = 1960 \text{ joules}$$

Thus, the total mechanical energy at the top is 1960 joules. As the ball falls, its potential energy converts into kinetic energy. At the bottom, the height is zero, meaning the potential energy is zero, and all the mechanical energy is now kinetic energy. The kinetic energy can be expressed as:

$$K = \frac{1}{2} mv^2$$

To find the final velocity just before impact, we can use the conservation of mechanical energy, which states:

$$K_{initial} + U_{initial} = K_{final} + U_{final}$$

Substituting the known values, we find that:

$$0 + 1960 = \frac{1}{2} mv_{final}^2 + 0$$

Solving for the final velocity gives:

$$v_{final} = \sqrt{\frac{2U_{initial}}{m}}$$

In this case, the mass cancels out, allowing us to find the final speed without knowing the mass of the ball. This leads to the equation:

$$v_{final} = \sqrt{2gh}$$

For a height of 100 meters, this results in:

$$v_{final} = \sqrt{2 \times 9.8 \times 100} \approx 44.3 \text{ m/s}$$

In a similar scenario where a ball is dropped from a height of 50 meters, the same principles apply. The initial potential energy is calculated as:

$$U_{initial} = mg(50)$$

Using the conservation of energy, we can again derive the final speed:

$$v_{final} = \sqrt{2gh}$$

Substituting the values gives:

$$v_{final} = \sqrt{2 \times 9.8 \times 50} \approx 31.3 \text{ m/s}$$

This demonstrates that the conservation of mechanical energy allows us to analyze the motion of objects efficiently, confirming that energy is conserved as it transforms between potential and kinetic forms. Understanding this principle is crucial for solving various physics problems involving energy transformations.

Launching Up To A Height

In this scenario, we analyze the motion of a 4-kilogram object launched vertically from the ground with an initial speed of \( v_0 = 40 \, \text{m/s} \). Our goal is to determine the maximum height \( y_{\text{max}} \) the object reaches before coming to a stop. To solve this, we apply the principle of conservation of energy, which states that the total mechanical energy in a closed system remains constant.

The energy conservation equation can be expressed as:

\[ K_{\text{initial}} + U_{\text{initial}} = K_{\text{final}} + U_{\text{final}} \]

At the ground level (where we define \( y = 0 \)), the initial kinetic energy \( K_{\text{initial}} \) is given by:

\[ K_{\text{initial}} = \frac{1}{2} m v_0^2 \]

Substituting the values, we have:

\[ K_{\text{initial}} = \frac{1}{2} \times 4 \, \text{kg} \times (40 \, \text{m/s})^2 = 3200 \, \text{J} \]

Since the object is at ground level, the initial gravitational potential energy \( U_{\text{initial}} \) is:

\[ U_{\text{initial}} = 0 \, \text{J} \]

At the maximum height, the final kinetic energy \( K_{\text{final}} \) is zero because the object momentarily stops. Thus, we have:

\[ K_{\text{final}} = 0 \, \text{J} \]

The final gravitational potential energy \( U_{\text{final}} \) at height \( y_{\text{max}} \) is given by:

\[ U_{\text{final}} = m g y_{\text{max}} \]

Setting up the energy conservation equation, we get:

\[ \frac{1}{2} m v_0^2 + 0 = 0 + m g y_{\text{max}} \]

We can simplify this equation by canceling the mass \( m \) from both sides:

\[ \frac{1}{2} v_0^2 = g y_{\text{max}} \]

Rearranging to solve for \( y_{\text{max}} \), we find:

\[ y_{\text{max}} = \frac{1}{2} \frac{v_0^2}{g} \]

Substituting the known values, where \( g \approx 9.8 \, \text{m/s}^2 \):

\[ y_{\text{max}} = \frac{1}{2} \frac{(40 \, \text{m/s})^2}{9.8 \, \text{m/s}^2} \approx 81.6 \, \text{m} \]

This calculation shows that the object reaches a maximum height of approximately 81.6 meters, which is equivalent to about 250 feet. This height assumes ideal conditions without air resistance. Understanding this concept of energy conservation is crucial in physics, as it allows us to predict the behavior of objects in motion under the influence of gravity.

Throwing An Object Downwards

In this problem, we analyze the motion of a 6-kilogram object thrown downward from an initial height of 20 meters, ultimately reaching a final speed of 30 meters per second just before impact with the ground. To solve for the initial speed, we apply the principle of conservation of energy, which states that the total mechanical energy (kinetic plus potential) remains constant in the absence of non-conservative forces.

We start by defining our energy terms. The initial kinetic energy (K_initial) is given by the formula:

K_initial = \frac{1}{2} m v_{initial}^2

where m is the mass of the object and v_initial is the initial velocity we want to find. The initial gravitational potential energy (U_initial) is calculated as:

U_initial = m g y_{initial}

where g is the acceleration due to gravity (approximately 9.8 m/s²) and y_initial is the height (20 meters).

At the final state, just before hitting the ground, the kinetic energy (K_final) is:

K_final = \frac{1}{2} m v_{final}^2

and the potential energy (U_final) is zero since the height is 0 meters.

Setting up the conservation of energy equation:

K_initial + U_initial = K_final + U_final

Substituting the expressions for kinetic and potential energy, we have:

\frac{1}{2} m v_{initial}^2 + m g y_{initial} = \frac{1}{2} m v_{final}^2

Since mass (m) appears in all terms, it can be canceled out, simplifying our equation to:

\frac{1}{2} v_{initial}^2 + g y_{initial} = \frac{1}{2} v_{final}^2

To eliminate fractions, we can multiply the entire equation by 2:

v_{initial}^2 + 2 g y_{initial} = v_{final}^2

Rearranging to solve for the initial velocity squared gives us:

v_{initial}^2 = v_{final}^2 - 2 g y_{initial}

Substituting the known values (v_final = -30 m/s, g = 9.8 m/s², and y_initial = 20 m), we find:

v_{initial}^2 = (-30)^2 - 2 \cdot 9.8 \cdot 20

Calculating this yields:

v_{initial}^2 = 900 - 392 = 508

Taking the square root gives us:

v_{initial} = \pm \sqrt{508} \approx \pm 22.5 \text{ m/s}

Since the object is thrown downward, we take the negative value, resulting in:

v_{initial} \approx -22.5 \text{ m/s}

This indicates that the object was thrown downward with an initial speed of approximately 22.5 meters per second, confirming that both upward and downward throws could yield the same final speed, but only the downward throw aligns with the problem's conditions.

Conservation Of Energy with Multiple Points

When launching a ball directly upwards, we can analyze its motion using the principles of energy conservation. Initially, the ball is at ground level (y = 0) and is launched with an unknown speed, which we will denote as \( v_a \). At a later point, the ball reaches a height of 30 meters with a speed of 20 meters per second. Eventually, it will reach its maximum height, where its speed will be 0.

To solve for the launch speed \( v_a \) and the maximum height \( y_{max} \), we can apply the conservation of mechanical energy. The total mechanical energy at any point is the sum of kinetic energy (KE) and gravitational potential energy (PE). The equations for kinetic and potential energy are given by:

KE = \( \frac{1}{2} mv^2 \) and PE = \( mgy \)

For part (a), we will consider the energy conservation from the initial point (A) at ground level to the point (B) where the ball is at 30 meters with a speed of 20 m/s. The energy conservation equation can be expressed as:

\( KE_A + PE_A = KE_B + PE_B \)

Since the ball starts from the ground, \( PE_A = 0 \). Thus, the equation simplifies to:

\( \frac{1}{2} mv_a^2 = \frac{1}{2} mv_b^2 + mg y_b \)

Here, \( v_b = 20 \, \text{m/s} \) and \( y_b = 30 \, \text{m} \). The mass \( m \) cancels out, leading to:

\( v_a^2 = v_b^2 + 2gy_b \)

Substituting the known values, we have:

\( v_a^2 = 20^2 + 2 \cdot 9.8 \cdot 30 \)

Calculating this gives:

\( v_a^2 = 400 + 588 = 988 \)

Taking the square root, we find:

\( v_a \approx 31.4 \, \text{m/s} \)

For part (b), we want to find the maximum height \( y_{max} \). We can use the energy conservation from the initial point (A) to the maximum height point (C). The equation becomes:

\( KE_A + PE_A = KE_C + PE_C \)

At maximum height, the kinetic energy \( KE_C = 0 \) because the speed is 0. Thus, we have:

\( \frac{1}{2} mv_a^2 = mgy_{max} \)

Again, the mass cancels out, leading to:

\( \frac{1}{2} v_a^2 = gy_{max} \)

Rearranging gives:

\( y_{max} = \frac{v_a^2}{2g} \)

Substituting \( v_a = 31.4 \, \text{m/s} \) into the equation:

\( y_{max} = \frac{(31.4)^2}{2 \cdot 9.8} \)

Calculating this yields:

\( y_{max} = \frac{985.96}{19.6} \approx 50.3 \, \text{m} \)

Thus, the maximum height the ball reaches is approximately 50.3 meters. This result is consistent with the ball still having speed at 30 meters, indicating it can ascend higher before coming to a stop.

Conservation Of Total Energy & Isolated Systems

Understanding the conservation of total energy is crucial in physics, particularly when analyzing isolated systems. Total energy encompasses all forms of energy within a system, including mechanical energy (kinetic and potential) and non-mechanical energy (thermal and others). For a system to maintain energy conservation, it must be isolated, meaning no external forces are acting on it. An isolated system only experiences internal forces, which are forces that occur between objects within the system.

To illustrate this concept, consider a scenario involving a box and a spring. If we define the system as just the box, the spring force acting on it is external, as it originates from outside the defined system. Consequently, the system is not isolated, leading to a situation where energy is not conserved. For example, if the initial kinetic energy of the box is 20 joules and the final kinetic energy after being launched by the spring is 30 joules, the total energy has changed, indicating that energy conservation does not hold.

Now, if we expand our system to include both the box and the spring, the forces acting between them become internal. The spring exerts a force on the box, and by Newton's third law, the box exerts an equal and opposite force on the spring. In this case, since all forces are internal, the system is isolated, allowing for energy conservation. Here, if the initial total energy (kinetic plus potential) is 30 joules and the final total energy remains 30 joules, energy conservation is upheld.

In summary, for energy to be conserved in a system, it must be isolated with only internal forces at play. This principle is fundamental in solving problems related to mechanical energy and understanding the dynamics of systems in physics.

Systems & Conservative vs. Non-Conservative Forces

Understanding the conservation of mechanical energy is crucial in physics, particularly when analyzing systems involving forces. A system is defined as a collection of objects, and the choice of what constitutes this system can significantly impact whether mechanical energy is conserved. For instance, consider a scenario where a spring pushes a box. If we analyze the box alone, we find that its kinetic energy changes from 20 joules to 30 joules after the spring releases it, indicating that mechanical energy is not conserved in this case. This is because the spring's influence was not included in our system.

However, if we expand our system to include both the box and the spring, we can account for all forms of energy. Initially, the total mechanical energy (kinetic plus potential) is 30 joules, and it remains 30 joules after the spring releases the box. This demonstrates that mechanical energy is conserved when the system is appropriately defined to include all relevant forces.

Mechanical energy is conserved when only conservative forces are acting on an object. Conservative forces, such as gravitational and spring forces, allow for energy to be transformed between potential and kinetic forms without loss. In contrast, nonconservative forces, like friction or applied forces, result in energy being added or removed from the system, leading to a decrease in total mechanical energy.

For example, when a block falls without air resistance, gravitational potential energy is converted into kinetic energy, illustrating energy transfer while conserving mechanical energy. Conversely, if a block is pushed and accelerates due to an applied force, energy is added to the system, indicating that mechanical energy is not conserved in this scenario.

Friction serves as another example of a nonconservative force. When a moving box slows down due to friction, kinetic energy is transformed into heat, resulting in a loss of mechanical energy. This highlights the fundamental difference between conservative and nonconservative forces: the former allows for reversible energy exchanges, while the latter leads to irreversible energy loss.

In summary, the conservation of mechanical energy is contingent upon the nature of the forces acting on a system. By carefully defining the system and recognizing the types of forces involved, one can determine whether mechanical energy is conserved and understand the energy transformations that occur within the system.

Here’s what students ask on this topic:

What is the principle of conservation of mechanical energy?

The principle of conservation of mechanical energy states that the total mechanical energy in an isolated system remains constant when only conservative forces act. This means that the sum of kinetic energy (K) and potential energy (U) does not change over time. Mathematically, it is expressed as:

$K_i+U_i=K_f+U_f$

where Ki and Ui are the initial kinetic and potential energies, and Kf and Uf are the final kinetic and potential energies. This principle is crucial for solving various physics problems efficiently.

How do you calculate the mechanical energy of a falling object?

To calculate the mechanical energy of a falling object, you need to consider both its kinetic energy (K) and potential energy (U). Initially, if the object is at rest, its kinetic energy is zero, and its potential energy is given by:

$U=mgh$

where m is the mass, g is the acceleration due to gravity, and h is the height. As the object falls, its potential energy decreases while its kinetic energy increases, given by:

$K=\frac{1}{2}m{v}^2$

where v is the velocity. The total mechanical energy remains constant, so:

$K_i+U_i=K_f+U_f$

What is an isolated system in the context of energy conservation?

An isolated system in the context of energy conservation is a system where no external forces are doing work on the system. This means that all forces acting within the system are internal. In such a system, the total energy, which includes both mechanical (kinetic and potential) and non-mechanical energies (like thermal energy), remains constant. For example, if you consider a box and a spring as an isolated system, the forces between the box and the spring are internal, and the total energy of the system is conserved.

How do conservative and nonconservative forces affect mechanical energy?

Conservative forces, such as gravity and spring forces, do not change the total mechanical energy of a system. They allow energy to be converted between kinetic and potential forms without any loss. For example, when a ball falls, its gravitational potential energy converts to kinetic energy. Nonconservative forces, like friction and applied forces, do change the total mechanical energy. They either add energy to the system or remove it, often converting mechanical energy into other forms like heat. This means that in the presence of nonconservative forces, the total mechanical energy is not conserved.

What is the difference between kinetic and potential energy in mechanical systems?

Kinetic energy (K) is the energy an object possesses due to its motion, given by the equation:

$K=\frac{1}{2}m{v}^2$

where m is the mass and v is the velocity. Potential energy (U) is the energy stored in an object due to its position or configuration. For example, gravitational potential energy is given by:

$U=mgh$

where h is the height above a reference point. In mechanical systems, these two forms of energy can convert into each other while the total mechanical energy remains constant in the absence of nonconservative forces.