Problem 81 - University Physics with Modern Physics

Question

A mass of \(1.00 \mathrm{~kg}\) attached to a spring with a spring constant of \(100 .\) N/m oscillates horizontally on a smooth frictionless table with an amplitude of \(0.500 \mathrm{~m} .\) When the mass is \(0.250 \mathrm{~m}\) away from equilibrium, determine: a) its total mechanical energy; b) the system's potential energy and the mass's kinetic energy; c) the mass's kinetic energy when it is at the equilibrium point. d) Suppose there was friction between the mass and the table so that the amplitude was cut in half after some time. By what factor has the mass's maximum kinetic energy changed? e) By what factor has the maximum potential energy changed?

Step-by-Step Solution

Answer

Answer: When the amplitude of the oscillating mass is halved, both the maximum kinetic energy and the maximum potential energy change by a factor of 1/4.

Step 1: a) Total Mechanical Energy

The total mechanical energy of the oscillating mass is the sum of its kinetic energy (KE) and potential energy (PE) due to the spring. In this case, the mechanical energy remains constant throughout the motion. To find the total mechanical energy, we first need to find the maximum potential energy when the mass is at its maximum amplitude, where the entire energy is potential energy and there's no kinetic energy. PE_max = (1/2) * k * A^2 Here, k is the spring constant, and A is the amplitude. PE_max = (1/2) * 100 * (0.500)^2 = 12.5 J Since the total mechanical energy is constant, we can say: E_total = PE_max = 12.5 J

Step 2: b) Potential Energy and Kinetic Energy

At a distance of 0.250 m from the equilibrium point, the potential energy (PE) and kinetic energy (KE) can be calculated using Hooke's Law and the conservation of energy. PE = (1/2) * k * x^2 Here, x is the displacement from the equilibrium point. PE = (1/2) * 100 * (0.250)^2 = 3.125 J As the mechanical energy is conserved, the sum of PE and KE at this point must be equal to the total mechanical energy. KE = E_total - PE = 12.5 - 3.125 = 9.375 J

Step 3: c) Kinetic Energy at Equilibrium Point

At the equilibrium point, the spring is at its natural length, and thus, there's no potential energy. Therefore, at this point, the entire mechanical energy is in the form of kinetic energy. KE_equilibrium = E_total = 12.5 J

Step 4: d) Maximum Kinetic Energy after Amplitude is Halved

If friction causes the amplitude to be halved after some time, we need to determine the new maximum potential energy and then find the maximum kinetic energy. New amplitude (A') = A / 2 = 0.500 / 2 = 0.250 m New maximum potential energy (PE_max') = (1/2) * k * (A')^2 = (1/2) * 100 * (0.250)^2 = 3.125 J The new total mechanical energy (E_total') = PE_max' = 3.125 J Now, at the maximum kinetic energy (occurring at the equilibrium point), there will be no potential energy. So, the new maximum kinetic energy is equal to the new total mechanical energy. KE_max' = E_total' = 3.125 J The factor by which the maximum kinetic energy has changed is: Factor = KE_max' / KE_equilibrium = 3.125 / 12.5 = 1 / 4 So, the maximum kinetic energy has changed by a factor of 1/4.

Step 5: e) Maximum Potential Energy Factor Change

We found the new maximum potential energy (PE_max') to be 3.125 J. We can now find the factor by which the maximum potential energy has changed: Factor = PE_max' / PE_max = 3.125 / 12.5 = 1 / 4 So, the maximum potential energy has also changed by a factor of 1/4.

Key Concepts

Mechanical Energy

Mechanical energy in a spring-mass system is the total energy comprising kinetic energy (energy due to motion) and potential energy (energy stored in the spring). This energy remains constant in an ideal, frictionless environment. In the given exercise, the mechanical energy is derived entirely from potential energy when the spring is at maximum compression or extension. Hence, the total mechanical energy is calculated as the potential energy when the spring is fully extended or compressed. In our example, with a spring constant ( \( k \)) of 100 N/m and an amplitude ( \( A \)) of 0.500 m:

• Potential Energy ( \( PE_{max} \)): \( \frac{1}{2} k A^2 = \frac{1}{2} \times 100 \times (0.500)^2 = 12.5 \) J

Thus, the mechanical energy of the system is 12.5 Joules at all times in the absence of external forces like friction.

Spring-Mass System

A spring-mass system involves an object attached to a spring that oscillates back and forth when displaced from its equilibrium position. When the system is frictionless and ideal, energy is transferred between kinetic and potential forms without loss. In our exercise, a 1.00 kg mass connected to a spring with a spring constant of 100 N/m illustrates this concept. The spring-mass system's behavior relies on Hooke's Law:

• Hooke's Law: Force exerted by a spring is proportional to the displacement: \( F = -kx \)
• Equilibrium Position: The point where the forces on the mass are balanced, and it experiences no net force.
• Amplitude: Maximum distance from equilibrium, showing the maximum stretch or compression of the spring.

These principles are vital for understanding how potential energy stored in the spring converts to kinetic energy as the mass moves, reversing phase between maximum potential energy and kinetic energy.

Kinetic Energy

Kinetic energy is the energy of motion. In the context of the spring-mass system, whenever the mass is moving, it possesses kinetic energy. The simple formula used to calculate kinetic energy is:

• \( KE = \frac{1}{2}mv^2 \)

However, in an oscillating spring-mass system without external forces, the kinetic energy can also be known from the conservation of mechanical energy. At the equilibrium position, kinetic energy is at its maximum, while potential energy is minimized. As derived in the step-by-step solution:

• Kinetic Energy at Equilibrium: 12.5 J when the entire mechanical energy converts to kinetic energy.
• Kinetic Energy at 0.250 m from Equilibrium: 9.375 J \( KE = E_{total} - PE\)

These values give insights into how energy transfer occurs and ensure energy conservation principles in the oscillation.

Potential Energy

Potential energy in a spring-mass system is the energy stored due to the position of the mass. It depends on both the spring constant and displacement from the equilibrium position. The formula used to calculate potential energy is:

• \( PE = \frac{1}{2}kx^2 \)

In the given problem, when the mass is at 0.250 m from equilibrium, the potential energy is calculated as:

• \( PE = \frac{1}{2} × 100 × (0.250)^2 = 3.125 \) J

Moreover, if the amplitude is halved due to some external force like friction, the potential energy is also reduced. This change reflects in further reducing mechanical energy, impacting both potential and kinetic energy. Here, for a new constrained amplitude of 0.250 m post-friction:

• New Potential Energy: \( 3.125 \) J, a reduction factor of 1/4 compared to the initial potential energy.

Potential energy plays a crucial role in balancing the system's energy conservation, transitioning energy forms during oscillation.