Problem 20 - University Physics with Modern Physics

Question

A 1.20-kg piece of cheese is placed on a vertical spring of negligible mass and force constant \(k =\) 1800 N/m that is compressed 15.0 cm. When the spring is released, how high does the cheese rise from this initial position? (The cheese and the springare \(not\) attached.)

Step-by-Step Solution

Answer

The cheese rises approximately 1.72 meters.

Step 1: Determine Potential Energy of the Spring

Calculate the potential energy stored in the compressed spring using the formula for elastic potential energy, which is \( U = \frac{1}{2} k x^2 \). Here, \( k = 1800 \) N/m and \( x = 0.15 \) m (converted from cm to m). So, the potential energy \( U \) becomes: \[ U = \frac{1}{2} \times 1800 \,\text{N/m} \times (0.15 \,\text{m})^2 = 20.25 \,\text{J}. \]

Step 2: Use Conservation of Energy

When the spring releases its energy, this potential energy is converted to kinetic energy and then to gravitational potential energy at the peak of the cheese's ascent. By conservation of energy, we have: \( \frac{1}{2} k x^2 = mgh \), where \( m = 1.20 \) kg, \( g = 9.81 \,\text{m/s}^2 \), and \( h \) is the maximum height reached. Therefore, \( 20.25 \,\text{J} = 1.20 \,\text{kg} \times 9.81 \,\text{m/s}^2 \times h \).

Step 3: Solve for Height

Rearrange the equation from Step 2 to solve for the height \( h \): \[ h = \frac{20.25 \,\text{J}}{1.20 \,\text{kg} \times 9.81 \,\text{m/s}^2} = \frac{20.25}{11.772} \approx 1.72 \,\text{m}. \] The cheese rises approximately 1.72 meters above its initial position on the spring.

Key Concepts

Elastic Potential Energy

Elastic potential energy is a type of energy stored in objects that can be deformed elastically, such as springs. When a spring is compressed or stretched from its natural position, it accumulates potential energy, waiting to be released when the force is removed. This potential energy is captured by the formula:

• \( U = \frac{1}{2} k x^2 \)

In this formula, \( U \) represents the elastic potential energy stored in the spring, \( k \) is the spring force constant, and \( x \) is the displacement of the spring from its equilibrium position.

In the given exercise, the spring is compressed by 0.15 meters, and the spring constant \( k \) is 1800 N/m. By plugging these values into the formula, we find that the stored energy in the spring is 20.25 Joules.

When the spring is released, this potential energy is transformed into kinetic and then gravitational potential energy, propelling the object placed on the spring upwards.

Gravitational Potential Energy

Gravitational potential energy is the energy an object possesses due to its position in a gravitational field. Essentially, it’s the energy stored because of an object's height above the ground. The higher an object is lifted, the more gravitational potential energy it has.

This energy is determined by the formula:

• \( U_g = mgh \)

Where \( U_g \) is the gravitational potential energy, \( m \) is the mass of the object, \( g \) is the acceleration due to gravity (9.81 m/s² on Earth), and \( h \) is the height of the object above the reference point.

In the problem with the piece of cheese and the spring, as the cheese rises, the elastic potential energy stored in the compressed spring is transformed into gravitational potential energy at the maximum height. This conservation of energy is crucial to determining how high the cheese will rise.

Spring Force Constant

The spring force constant, denoted by \( k \), is a measure of a spring's stiffness. It indicates how much force is needed to compress or stretch a spring by a certain distance. A higher spring constant means a stiffer spring, requiring more force for the same amount of displacement.

In the spring force equation, Hooke’s Law, it is expressed as:

• \( F = kx \)

Here, \( F \) is the force applied, \( k \) is the spring constant, and \( x \) is the displacement from the spring's relaxed position.

For the exercise, the spring used has a constant \( k = 1800 \) N/m, indicating it is relatively firm. This stiffness contributes to the amount of potential energy that can be stored when the spring is compressed, thus impacting how high the cheese will rise once this energy is released.