Problem 80 - University Physics with Modern Physics

Question

A 0.100 -kg ball is dropped from a height of \(1.00 \mathrm{~m}\) and lands on a light (approximately massless) cup mounted on top of a light, vertical spring initially at its equilibrium position. The maximum compression of the spring is to be \(10.0 \mathrm{~cm}\). a) What is the required spring constant of the spring? b) Suppose you ignore the change in the gravitational energy of the ball during the 10 -cm compression. What is the percentage difference between the calculated spring constant for this case and the answer obtained in part (a)?

Step-by-Step Solution

Answer

Answer: The spring constant of the light, vertical spring is 196.2 N/m. The percentage difference between the calculated spring constants considering and ignoring the change in gravitational energy during the 10-cm compression is approximately 0.10%.

Step 1: Noting the given information

We have the following information given in the problem: 1. Mass of the ball (m) = 0.100 kg 2. Height from which ball is dropped (h) = 1.00 m 3. Maximum compression of the spring (x) = 10.0 cm = 0.1 m (converted to meters for consistency)

Step 2: Determine the initial and final energies

At the initial moment before dropping the ball, it only has gravitational potential energy: Initial total energy (E_initial) = Gravitational potential energy = m * g * h At the final moment when the spring is maximally compressed, the ball has both gravitational potential energy and spring potential energy: Final total energy (E_final) = Gravitational potential energy + Spring potential energy = (m * g * (h - x)) + (1/2 * k * x^2) Here k is the spring constant, which we need to find.

Step 3: Apply conservation of mechanical energy

Since the mechanical energy is conserved, we can equate the initial and final energies: E_initial = E_final m * g * h = (m * g * (h - x)) + (1/2 * k * x^2)

Step 4: Solve for the spring constant, k

We will now isolate k from the equation above and solve for it. m * g * h - m * g * (h - x) = (1/2 * k * x^2) k = (2 * (m * g * h - m * g * (h - x))) / x^2 Now substituting the given values to calculate k: k = (2 * (0.100 kg * 9.81 m/s^2 * 1.00 m - 0.100 kg * 9.81 m/s^2 * (1.00 m - 0.1 m))) / (0.1 m)^2 k = 196.2 N/m So, the required spring constant is k = 196.2 N/m.

Step 5: Calculate the spring constant without considering the change in gravitational potential energy

Now, if we ignore the change in gravitational potential energy when the spring is compressed, we have to only consider the spring potential energy at the final moment: E_final' = Spring potential energy = (1/2 * k * x^2) E_initial = E_final' m * g * h = (1/2 * k * x^2) k' = (2 * (m * g * h)) / x^2 Now substituting the given values to calculate k': k' = (2 * (0.100 kg * 9.81 m/s^2 * 1.00 m)) / (0.1 m)^2 k' = 196.0 N/m So, the calculated spring constant without considering the change in gravitational potential energy is k' = 196.0 N/m.

Step 6: Calculate the percentage difference

The percentage difference between the calculated spring constants k and k' can be calculated as follows: Percentage Difference = 100 * (|k - k'|) / k Percentage Difference = 100 * (|196.2 N/m - 196.0 N/m|) / 196.2 N/m Percentage Difference ≈ 0.10% The percentage difference between the calculated spring constants is approximately 0.10%.

Key Concepts

Potential Energy

Potential energy is energy that is stored in an object due to its position or configuration. In this exercise, when the ball is held at a height before being dropped, it has gravitational potential energy. This type of potential energy depends on the height from a reference point—in this case, the ground—and the mass of the object, as well as the gravitational acceleration. This can be calculated using the formula: \[ \text{Potential Energy} = m \cdot g \cdot h \] where:

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

Understanding potential energy is crucial in determining how much energy is available to be converted into other forms, such as kinetic energy or the energy stored in a spring.

Spring Constant

The spring constant, often represented by \( k \), is a measure of the stiffness of a spring. It is defined as the force required to compress or extend the spring by one unit length. In this exercise, after the ball is dropped and lands on the spring, the spring's compression is directly related to the spring constant.
The spring potential energy, or elastic potential energy, is calculated using the formula: \[ \text{Spring Potential Energy} = \frac{1}{2} \cdot k \cdot x^2 \] where:

• \( k \) is the spring constant,
• \( x \) is the displacement from the equilibrium position (or compression in this case).

The spring constant is crucial in this calculation because it determines how much the spring will compress under a given force (in this case, the weight of the ball). In the solution, we derived \( k \) to be \( 196.2 \, \text{N/m} \), indicating the required stiffness for the spring to only compress by the specified amount (10 cm).

Mechanical Energy Conservation

The principle of mechanical energy conservation states that in a closed system with only conservative forces acting, the total mechanical energy remains constant. This concept is vital in understanding how energy is transferred through different forms in this exercise.
Initially, the ball has gravitational potential energy while being held at the initial height. Upon release, as it descends, this potential energy is converted to kinetic energy, as well as eventually into spring potential energy when it compresses the spring. The equation used to express this conservation of energy is: \[ \text{Initial Energy} = \text{Final Energy} \] That is, \[ m \cdot g \cdot h = (m \cdot g \cdot (h - x)) + \frac{1}{2} \cdot k \cdot x^2 \] This reflects how the energy at the start (only gravitational) equals the sum of gravitational and spring energy at maximum compression. By following this rule, we can accurately find unknowns like the spring constant which is necessary for such transformations.

Gravitational Potential Energy

Gravitational potential energy is the energy an object possesses because of its position in a gravitational field. When the ball is dropped in the exercise, it possesses gravitational potential energy given by the formula: \[ \text{Gravitational Potential Energy} = m \cdot g \cdot h \] This quantity changes as the ball falls and is crucial for understanding energy conversion in this setup.
By ignoring the change in gravitational potential energy, as done in part of the exercise, the spring constant result slightly changes. It shows how dependent calculations can be on whether or not we account for such forms of energy fully—resulting in a different value (\( 196.0 \, \text{N/m} \) instead of \( 196.2 \, \text{N/m} \)) and causing a percentage difference of about 0.10%. The lesson here is how each component of potential energy affects the results, stressing the importance of considering every factor in physical predictions.