More Conservation of Energy Problems: Videos & Practice Problems

In a system with two blocks connected by a light rope over a pulley, the conservation of energy principle is applied to find the final speed of the blocks. The potential energy of the falling block is converted into kinetic energy, accounting for work done against friction. The equation derived is $\sqrt{\frac{4gh(m2-m1)}{2m1+2m2+m3}}$. The final speed calculated is approximately 6.47 m/s.

Speed of blocks on a pulley (Atwood's Machine)

In this problem, we explore the dynamics of Atwood's machine, which consists of two blocks connected by a light string over a pulley. The blocks have masses of 3 kg and 5 kg, while the pulley is modeled as a solid cylinder with a mass of 4 kg and a radius of 8 m. The system is released from rest, with the heavier block (5 kg) starting at a height of 5 m above the ground. The goal is to determine the speed of the heavier block just before it hits the ground and the angular speed of the pulley at that moment.

To solve this, we apply the principle of conservation of energy, which states that the total mechanical energy in a closed system remains constant if only conservative forces are acting. The initial potential energy of the system is converted into kinetic energy as the blocks move and the pulley rotates. The relevant equations include:

1. **Potential Energy (PE)**: \( U = mgh \), where \( m \) is mass, \( g \) is the acceleration due to gravity (approximately 10 m/s²), and \( h \) is height.

2. **Kinetic Energy (KE)**: For linear motion, \( KE = \frac{1}{2} mv^2 \), and for rotational motion, \( KE = \frac{1}{2} I \omega^2 \), where \( I \) is the moment of inertia and \( \omega \) is the angular velocity.

3. **Moment of Inertia for a solid cylinder**: \( I = \frac{1}{2} m r^2 \), where \( r \) is the radius of the cylinder.

Initially, the system has potential energy due to the height of the 5 kg block, while the kinetic energy is zero since the system starts from rest. As the blocks move, the potential energy of the descending block is converted into kinetic energy of both blocks and the rotational kinetic energy of the pulley. The conservation of energy equation can be expressed as:

Initial Potential Energy = Final Kinetic Energy

Thus, we have:

\( m_2gh_{2i} = \frac{1}{2} m_1 v^2 + \frac{1}{2} m_2 v^2 + \frac{1}{2} I \omega^2 \)

Substituting \( I \) and \( \omega \) in terms of \( v \) (using \( \omega = \frac{v}{r} \)), we can simplify the equation to find the final speed \( v \) of the blocks just before the heavier block hits the ground:

\( v_{final} = \sqrt{\frac{4gh(m_2 - m_1)}{2m_1 + 2m_2 + m_3}} \)

After substituting the known values (with \( g = 10 \, \text{m/s}^2 \), \( h = 5 \, \text{m} \), \( m_1 = 3 \, \text{kg} \), \( m_2 = 5 \, \text{kg} \), and \( m_3 = 4 \, \text{kg} \)), we find that the final speed \( v \) is approximately 4.5 m/s.

For part b, to find the angular speed \( \omega \) of the pulley, we use the relationship \( \omega = \frac{v}{r} \). Substituting the calculated speed and the radius of the pulley, we can determine the angular speed just before the block hits the ground.

This problem illustrates the application of conservation of energy in a rotational system, highlighting the interplay between linear and rotational motion in a connected system.

Blocks on a rough table and a pulley

In this problem, we analyze a system consisting of two blocks connected by a light rope that passes through a pulley. The pulley is modeled as a solid cylinder, which has a moment of inertia given by the formula \( I = \frac{1}{2} m r^2 \). Here, we define the masses of the blocks as \( m_1 = 4 \, \text{kg} \) (horizontal block), \( m_2 = 6 \, \text{kg} \) (vertical block), and \( m_3 = 10 \, \text{kg} \) (pulley), with a radius \( r = 2 \, \text{m} \). The coefficient of friction between the horizontal block and the surface is \( \mu = 0.5 \).

The system is released from rest, meaning the initial velocity \( v_{2, \text{initial}} = 0 \). The vertical block starts at a height of \( h_{2, \text{initial}} = 8 \, \text{m} \) and we want to find its speed just before it hits the floor, where \( h_{2, \text{final}} = 0 \). Since the blocks are connected, their velocities are the same, denoted as \( v \).

To solve for the final velocity, we apply the principle of conservation of energy, which states that the total mechanical energy in the system remains constant if only conservative forces are acting. The equation can be expressed as:

\[ K_{\text{initial}} + U_{\text{initial}} + W_{\text{non-conservative}} = K_{\text{final}} + U_{\text{final}} \]

Initially, the kinetic energy \( K_{\text{initial}} = 0 \) since the system starts from rest. The only potential energy contributing to the initial energy is from the vertical block:

\[ U_{\text{initial}} = m_2 g h_{2, \text{initial}} \]

As the system moves, work is done against friction, which is calculated as:

\[ W_{\text{friction}} = -\mu m_1 g d \]

At the final state, the kinetic energy consists of the kinetic energy of both blocks and the rotational kinetic energy of the pulley:

\[ K_{\text{final}} = \frac{1}{2} m_1 v^2 + \frac{1}{2} m_2 v^2 + \frac{1}{2} I \omega^2 \]

Substituting the moment of inertia and the relationship between linear and angular velocity \( \omega = \frac{v}{r} \), we have:

\[ K_{\text{final}} = \frac{1}{2} m_1 v^2 + \frac{1}{2} m_2 v^2 + \frac{1}{2} \left(\frac{1}{2} m_3 r^2\right) \left(\frac{v}{r}\right)^2 \]

By simplifying, we can express the total kinetic energy in terms of \( v \). The potential energy at the final state is zero since the vertical block is at ground level.

Combining all these components, we arrive at the equation:

\[ m_2 g h_{2, \text{initial}} - \mu m_1 g d = \frac{1}{2} m_1 v^2 + \frac{1}{2} m_2 v^2 + \frac{1}{4} m_3 v^2 \]

Recognizing that the distance \( d \) moved by the horizontal block is equal to the height \( h \) dropped by the vertical block, we can substitute \( d \) with \( h \). After rearranging and factoring out common terms, we can isolate \( v \) to find:

\[ v = \sqrt{\frac{4gh(m_2 - \mu m_1)}{2m_1 + 2m_2 + m_3}} \]

Substituting the known values into this equation yields a final speed of approximately \( 6.47 \, \text{m/s} \) for the vertical block just before it hits the floor. This type of problem illustrates the application of energy conservation principles in systems involving pulleys and friction.

Speed of a yo-yo

In this scenario, we analyze a simple yo-yo, which is modeled as a solid disc, to determine its linear and angular speeds after it falls a certain distance. The yo-yo has a mass of 0.1 kg and an inner radius of 0.02 m. It is released from rest, meaning its initial velocity is zero, and it falls a height of 0.5 m while unwinding a light string around its cylindrical shaft.

To solve for the final linear speed (\(v_f\)) and angular speed (\(\omega_f\)), we apply the principle of conservation of energy. Initially, the yo-yo possesses gravitational potential energy given by the formula:

\(PE = mgh\)

where \(m\) is the mass, \(g\) is the acceleration due to gravity (approximately \(9.81 \, \text{m/s}^2\)), and \(h\) is the height (0.5 m). As the yo-yo falls, this potential energy is converted into both translational kinetic energy and rotational kinetic energy:

\(KE_{trans} = \frac{1}{2} mv_f^2\)

\(KE_{rot} = \frac{1}{2} I \omega_f^2\)

For a solid disc, the moment of inertia (\(I\)) is given by:

\(I = \frac{1}{2} m r^2\)

where \(r\) is the radius of the yo-yo. Additionally, the relationship between linear speed and angular speed is expressed as:

\(v_f = r \omega_f\)

By substituting \(I\) and \(\omega_f\) in terms of \(v_f\) into the energy conservation equation:

\(mgh = \frac{1}{2} mv_f^2 + \frac{1}{2} \left(\frac{1}{2} m r^2\right) \left(\frac{v_f}{r}\right)^2\)

After simplifying and canceling out the mass \(m\) and radius \(r\), we arrive at:

\(4gh = 3v_f^2\)

From this, we can solve for the final linear speed:

\(v_f = \sqrt{\frac{4}{3}gh}\)

This result indicates that the final linear speed depends solely on the height from which the yo-yo is released and the acceleration due to gravity, rather than its mass or radius. The coefficient \( \frac{4}{3} \) arises from the moment of inertia of the solid disc.

For the final angular speed, we can use the relationship between linear and angular speeds:

\(\omega_f = \frac{v_f}{r}\)

Substituting \(v_f\) into this equation gives:

\(\omega_f = \frac{1}{r} \sqrt{\frac{4}{3}gh}\)

This analysis highlights the differences between linear and rotational motion. While both types of motion share similar forms in their equations, the coefficients differ due to the distribution of mass and the nature of the motion. In this case, the yo-yo falls slower than a block dropped from the same height because some of its potential energy is converted into rotational kinetic energy, illustrating the interplay between linear and rotational dynamics.

A small 10-kg object is connected to the right end of a thin rod of length 4 m and mass 5 kg. The rod is free to rotate about a fixed perpendicular axis on its left end, as shown below. The rod is initially held at rest, horizontally. When the rod is released, it falls, rotating about its axis, similar to a pendulum. What is the speed at the rod's center of mass when the rod is vertical? BONUS:What is object's speed when the rod is vertical?

• A. 4.4 m/s
• B. 4.9 m/s
• C. 5.6 m/s
• D. 6.1 m/s

Answer: (A)

Here’s what students ask on this topic:

How do you apply the conservation of energy principle to a system with two blocks and a pulley?

To apply the conservation of energy principle to a system with two blocks and a pulley, you need to account for the potential energy, kinetic energy, and work done by non-conservative forces like friction. The total mechanical energy of the system remains constant if only conservative forces are acting. The equation used is:

$K_{\mathrm{initial}}+U_{\mathrm{initial}}+W_{\mathrm{non-conservative}}=K_{\mathrm{final}}+U_{\mathrm{final}}$

Here, Kinitial and Kfinal are the initial and final kinetic energies, Uinitial and Ufinal are the initial and final potential energies, and Wnon-conservative is the work done by non-conservative forces.

What is the role of friction in conservation of energy problems involving pulleys?

In conservation of energy problems involving pulleys, friction plays a crucial role as a non-conservative force. It does work against the motion of the blocks, converting some of the mechanical energy into thermal energy, which is not recoverable. The work done by friction is calculated using:

$W{\mathrm{friction}}_{}=\; -\mu Nd$

where μ is the coefficient of friction, N is the normal force, and d is the distance over which friction acts. This work is subtracted from the total mechanical energy of the system.

How do you calculate the final speed of blocks in a pulley system using energy conservation?

To calculate the final speed of blocks in a pulley system using energy conservation, you set up the energy conservation equation:

$K_{\mathrm{initial}}+U_{\mathrm{initial}}+W_{\mathrm{non-conservative}}=K_{\mathrm{final}}+U_{\mathrm{final}}$

For a system with two blocks and a pulley, the equation becomes:

$4\mathrm{gh}\left(m_2-m_1\right)=2m_1+2m_2+m_3$

Solving for the final speed v_final:

$v$_final = 4gh(m2 - m1)2m1 + 2m2 + m3

Plug in the given values to find the final speed.

What is the significance of the moment of inertia in pulley problems?

The moment of inertia is significant in pulley problems because it accounts for the rotational inertia of the pulley. It affects the system's total kinetic energy. For a solid cylinder pulley, the moment of inertia I is given by:

$I=\frac{1}{2}m{R}^2$

where m is the mass of the pulley and R is its radius. The rotational kinetic energy of the pulley is:

$\frac{1}{2}I{\omega }^2$

where ω is the angular velocity. This term must be included in the total kinetic energy of the system.

How do you handle potential energy changes in a pulley system with blocks?

In a pulley system with blocks, potential energy changes are handled by considering the height changes of the blocks. The potential energy U is given by:

$U=mgh$

where m is the mass, g is the acceleration due to gravity, and h is the height. For a block moving vertically, the change in potential energy is:

$\mathrm{\Delta U}=mg\mathrm{\Delta h}$

where Δh is the change in height. Only the block that changes height contributes to the potential energy change in the system.