Problem 24 - University Physics with Modern Physics

Question

{~A} 100 \cdot \mathrm{g}\( block hangs from a spring with \)k=5.00 \mathrm{~N} / \mathrm{m}\( At \)t=0 \mathrm{~s},\( the block is \)20.0 \mathrm{~cm}\( below the equilibrium posi. tion and moving upward with a speed of \)200, \mathrm{~cm} / \mathrm{s}\(. What is the block's speed when the displacement from equilibrium is \)30.0 \mathrm{~cm} ?$

Step-by-Step Solution

Answer

Answer: The block's speed when the displacement from equilibrium is 30 cm (0.3 m) is approximately 1.22 m/s.

Step 1: 1. Determine the initial conditions

At t=0 s, the block is 20 cm below the equilibrium position and moving upwards with an initial speed of 200 cm/s.

Step 2: 2. Convert units to SI

To facilitate calculations, let's convert all given measurements to standard units. Initial displacement (x1): 20 cm = 0.2 m Initial speed (v1): 200 cm/s = 2 m/s Spring constant (k): 5 N/m Final displacement (x2): 30 cm = 0.3 m

Step 3: 3. Calculate initial potential and kinetic energy

Next, calculate the potential energy stored in the spring (Us) and the initial kinetic energy (K1). Us = 0.5 * k * x1^2 = 0.5 * 5 * (0.2)^2 = 0.1 J K1 = 0.5 * m * v1^2, we need to find the mass (m) of the block. Given that weight = mg, we have m = W/g = (100*10^-3 kg) = 0.1 kg K1 = 0.5 * 0.1 * (2)^2 = 0.2 J

Step 4: 4. Apply conservation of energy

The total energy of the system at the initial position will be equal to the total energy at the final position (x2): E_total = Us + K1 = 0.1 J + 0.2 J = 0.3 J

Step 5: 5. Calculate final potential energy and kinetic energy

At x2, the potential energy stored in the spring is: Us2 = 0.5 * k * x2^2 = 0.5 * 5 * (0.3)^2 = 0.225 J Since conservation of energy holds, the final kinetic energy (K2) is: K2 = E_total - Us2 = 0.3 J - 0.225 J = 0.075 J

Step 6: 6. Find the final speed of the block

Now that we know the final kinetic energy, we can use it to find the final speed of the block (v2): K2 = 0.5 * m * v2^2 0.075 J = 0.5 * 0.1 * v2^2 Solve for v2: v2 = sqrt(0.075*2/0.1) ≈ 1.22 m/s The block's speed when the displacement from equilibrium is 30 cm (0.3 m) is approximately 1.22 m/s.

Key Concepts

Conservation of Energy

Understanding the principle of conservation of energy is essential when solving oscillation problems. This principle states that within a closed system, energy can neither be created nor destroyed; it can only be transformed from one form to another. In the exercise, a block attached to a spring converts energy between kinetic energy, when the block is moving, and potential energy stored in the spring, when it is stretched or compressed.

During the motion of the block, we expect that the total mechanical energy remains constant if there is no energy lost to external forces such as friction. Therefore, our calculations assume that the initial total energy, which is the sum of the initial kinetic energy and the spring potential energy, will equal the total energy at any subsequent time. This concept guides us to write the energy conservation equation and helps us solve for the unknown variables like the final speed of the block.

Simple Harmonic Motion

Simple harmonic motion (SHM) describes the motion of an object that experiences a restoring force proportional to its displacement from the equilibrium position. In the context of our exercise, when the block attached to the spring is displaced and then released, it exhibits SHM. One characteristic of SHM is that the object moves back and forth about the equilibrium position in a periodic fashion.

Mathematically, SHM can be described using the second law of motion, where the mass times the acceleration of the oscillating object is equal to the negative product of the spring constant and the displacement. This concept helps us understand that as the block moves towards or away from the equilibrium position, it's exchanging spring potential energy and kinetic energy within a predictable, sinusoidal pattern.

Spring Potential Energy

The concept of spring potential energy is fundamental in solving our block-and-spring system problem. The potential energy in a spring, also known as elastic potential energy, is energy stored as a result of deformation of an elastic object, such as the stretching or compressing of the spring in our scenario. This energy can be calculated using the formula:
U_s = 0.5 * k * x^2,
where U_s is the spring potential energy, k is the spring constant, and x is the displacement from the equilibrium position.

In equations, we capture how energy is stored in the spring and then released as kinetic energy when the spring returns to its natural length. In the exercise, we calculated the initial and final potential energy of the spring to determine the change in kinetic energy as the block moves from one position to another.

Kinetic Energy Calculation

The kinetic energy calculation is another critical component of solving oscillation problems. Kinetic energy refers to the energy an object possesses due to its motion. For a mass m, moving with a velocity v, the kinetic energy (K) is given by the formula:
K = 0.5 * m * v^2.

In the aforementioned problem, we use this formula to calculate both the initial and final kinetic energy of the block as it oscillates. By applying the conservation of energy, we use the initial kinetic energy, along with the potential energy of the spring, to find the kinetic energy at another point in the motion. Solving for the final velocity requires rearranging this equation to isolate the velocity variable after determining kinetic energy at the final position.

By following these steps, we were able to deduce the speed of the block at a 30 cm displacement from the equilibrium, using its relationship with kinetic energy.