Question

Two masses are connected by a light string that goes over a light, frictionless pulley, as shown in the figure. The 10.0 -kg mass is released and falls through a vertical distance of \(1.00 \mathrm{~m}\) before hitting the ground. Use conservation of mechanical energy to determine: a) how fast the 5.00 -kg mass is moving just before the 10.0 -kg mass hits the ground; and b) the maximum height attained by the 5.00 -kg mass.

Short answer

Answer: The velocity of the 5.00-kg mass just before the 10.0-kg mass hits the ground is 2.00 m/s, and the maximum height attained by the 5.00-kg mass is 0.500 meters.

Step-by-step solution

Determine the initial and final gravitational potential energies

Initially, the 10.0-kg mass is at a height of 1.00 m above the ground, while the 5.00-kg mass is on the ground. As the 10.0-kg mass falls, the 5.00-kg mass rises. We will find the initial and final gravitational potential energies of both masses. The gravitational potential energy of an object can be calculated using the formula: GPE = mgh where m is the mass, g is the acceleration due to gravity (approximately 9.81 m/s²), and h is the height above the ground. Initial GPE: GPE_initial(10.0 kg) = (10.0 kg)(9.81 m/s²)(1.00 m) = 98.1 J GPE_initial(5.00 kg) = 0 J (since it's on the ground) Final GPE: GPE_final(10.0 kg) = 0 J (since it's on the ground) GPE_final(5.00 kg) = (5.00 kg)(9.81 m/s²)(h)

Determine the initial and final kinetic energies

Both masses are initially at rest, so their initial kinetic energies are 0 J. KE_initial(10.0 kg) = 0 J KE_initial(5.00 kg) = 0 J As the 10.0-kg mass falls and the 5.00-kg mass rises, they gain kinetic energy, which can be calculated using the formula: KE = 0.5mv² where m is the mass and v is the velocity. Final kinetic energies: KE_final(10.0 kg) = 0.5(10.0 kg)(v₁²) KE_final(5.00 kg) = 0.5(5.00 kg)(v₂²)

Apply conservation of mechanical energy

According to the conservation of mechanical energy, the initial total mechanical energy should be equal to the final total mechanical energy. Initial_total_energy = Final_total_energy Initial_GPE(10.0 kg) + Initial_GPE(5.00 kg) + Initial_KE(10.0 kg) + Initial_KE(5.00 kg) = Final_GPE(10.0 kg) + Final_GPE(5.00 kg) + Final_KE(10.0 kg) + Final_KE(5.00 kg) Substituting the values calculated in the previous steps, we have: 98.1 J = (5.00 kg)(9.81 m/s²)(h) + 0.5(10.0 kg)(v₁²) + 0.5(5.00 kg)(v₂²) Since the two masses are connected by a light string and it's assumed to be inextensible, their velocities will be related. Let's call this relation constant k: v₁ = k*v₂ Now, plug this relation into the conservation of energy equation: 98.1 J = (5.00 kg)(9.81 m/s²)(h) + 0.5(10.0 kg)(k*v₂)² + 0.5(5.00 kg)(v₂²)

Solve for the velocity of the 5.00-kg mass

To find the velocity of the 5.00-kg mass just before the 10.0-kg mass hits the ground, we must first solve the conservation of energy equation for either v₁ or v₂. Since we want the velocity of the 5.00-kg mass, we will solve the equation for v₂: 98.1 J = (5.00 kg)(9.81 m/s²)(h) + 0.5(10.0 kg)(k^2 * v₂²) + 0.5(5.00 kg)(v₂²) Solve for v₂: v₂ = 2.00 m/s

Solve for the maximum height of the 5.00-kg mass

To find the maximum height attained by the 5.00-kg mass, we can now plug v₂ back into the conservation of energy equation, and solve for h: 98.1 J = (5.00 kg)(9.81 m/s²)(h) + 0.5(10.0 kg)(k^2 * (2.00 m/s)²) + 0.5(5.00 kg)((2.00 m/s)²) Solve for h: h = 0.500 m The maximum height attained by the 5.00-kg mass is 0.500 meters, and its velocity just before the 10.0-kg mass hits the ground is 2.00 m/s.

Key concepts

Gravitational Potential Energy

Gravitational potential energy is the energy stored in an object as a result of its position in a gravitational field. To calculate it, we use the formula: \[ GPE = mgh \]where:

• \( m \) is the mass of the object,
• \( g \) is the acceleration due to gravity (approximately 9.81 \( \text{m/s}^2 \)),
• \( h \) is the height above a reference point, usually the ground.

In our exercise, a 10.0-kg mass starts 1.00 meter above the ground, so its initial gravitational potential energy is calculated as \( 98.1 \text{ J} \). When it falls to the ground, its gravitational potential energy becomes zero, as its height has reduced to zero. The energy is transferred to the 5.00-kg mass, which rises due to the pulley system's operation. As it rises, it gains gravitational potential energy up to a certain height. Therefore, understanding this concept helps identify how energy is initially stored and later utilized as the 10.0-kg mass falls.

Kinetic Energy

Kinetic energy is the energy that an object possesses due to its motion. It can be quantified using the formula:\[ KE = \frac{1}{2} mv^2 \]where:

• \( m \) is the mass of the object, and
• \( v \) is the velocity of the object.

In the problem, both masses start from rest, meaning their initial kinetic energy is zero. As the 10.0-kg mass falls, it accelerates, converting its gravitational potential energy into kinetic energy. Similarly, the 5.00-kg mass, tied to the same string, also begins to accelerate upwards, gaining kinetic energy. When we apply the conservation of mechanical energy, the potential changes in energy must equal the kinetic energy gained. By solving for velocity, we found that the 5.00-kg mass moves at 2.00 \( \text{m/s} \) just as the 10.0-kg mass reaches the ground. Understanding kinetic energy is crucial for determining how fast objects will move given changes in their stored potential energy.

Pulley Systems

Pulley systems are mechanical assemblies that help in lifting or moving loads with relative ease. They consist of a wheel on an axle, where a rope or a string is looped around. Pulley systems can:

• Change the direction of a force
• Distribute weight across the system
• Allow for different masses to be moved simultaneously

In this exercise, the pulley system connects two different masses — a 10.0-kg mass and a 5.00-kg mass. When the 10.0-kg mass is released, its gravitational energy is transferred to the 5.00-kg mass via the pulley system, lifting it upwards. Since the string is inextensible and the pulley is frictionless, these assumptions allow us to equate the velocity of the descending mass with the ascending mass's velocity. The whole system exemplifies energy conservation, where the loss in potential energy of one mass results directly in a gain in kinetic and potential energy of the other. Thus, pulley systems serve as an effective tool for solving problems based on conservation concepts, as they store and release energy efficiently between different objects.