Problem 91 - College Physics

Question

\(\bullet\) \(\bullet\) A 250 g object on a frictionless, horizontal lab table is pushed against a spring of force constant 35 \(\mathrm{N} / \mathrm{cm}\) and then released. Just before the object is released, the spring is com- pressed 12.0 \(\mathrm{cm} .\) How fast is the object moving when it has gained half of the spring's original stored energy?

Step-by-Step Solution

Answer

The object moves at approximately 10.04 m/s.

Step 1: Define the Problem

We have a 250 g object and a spring with a force constant of 35 N/cm. The spring is compressed by 12.0 cm initially. We need to calculate the speed of the object when it has gained half of the spring's originally stored energy.

Step 2: Convert Units

Convert the force constant and mass into standard units. The force constant is 35 N/cm, which equals 3500 N/m. The mass of the object is 250 g, which is 0.250 kg.

Step 3: Calculate the Initial Potential Energy

The initial potential energy stored in the spring is given by \( U_i = \frac{1}{2} k x^2 \). Substituting for \( k = 3500 \text{ N/m} \) and \( x = 0.12 \text{ m} \), we get \[ U_i = \frac{1}{2} \times 3500 \times (0.12)^2 = 25.2 \text{ J}. \]

Step 4: Determine the Energy to Be Converted to Kinetic Energy

The object has gained half of the spring's initial energy as kinetic energy, therefore \( KE = \frac{1}{2} U_i = \frac{1}{2} \times 25.2 = 12.6 \text{ J} \).

Step 5: Use Kinetic Energy Formula

The kinetic energy is given by \( KE = \frac{1}{2} mv^2 \). Solving for \( v \), we have \[ \frac{1}{2} \times 0.250 \times v^2 = 12.6 \].

Step 6: Solve for Velocity

Rearrange the equation to find \( v^2 = \frac{2 \times 12.6}{0.250} \). Therefore, \( v^2 = 100.8 \), and \( v = \sqrt{100.8} \). Thus, the velocity \( v \approx 10.04 \text{ m/s} \).

Key Concepts

Potential Energy

Potential energy is the stored energy of an object due to its position or configuration. When we talk about a compressed spring, it holds energy called elastic potential energy. This form of energy is ready to do work when the spring is released to move back to its original shape.

In our example, the spring is compressed by 12.0 cm, storing potential energy. The potential energy in a spring is calculated using the formula \[ U = \frac{1}{2} k x^2 \] Where:

• \( U \) represents the potential energy,
• \( k \) is the spring constant, reflecting the stiffness of the spring,
• \( x \) is the distance the spring is compressed or stretched from its natural length.

This calculation shows how potential energy is converted into kinetic energy when the spring returns to its original length.

Kinetic Energy

Kinetic energy is the energy an object possesses due to its motion. In simple terms, if an object is moving, it has kinetic energy. When an object is released from a compressed spring, the stored potential energy in the spring is transformed into the kinetic energy of the object.

The formula for kinetic energy is:\[ KE = \frac{1}{2} mv^2 \] Where:

• \( KE \) is the kinetic energy
• \( m \) is the mass of the object
• \( v \) is the velocity of the object

In the scenario provided, half of the spring's initial potential energy is converted into the kinetic energy of the object. So, if we know the mass and the amount of kinetic energy, we can solve for the speed of the object as it moves. This is an application of energy conservation, showcasing how energy transforms from one type to another.

Spring Constant

The spring constant is an essential concept when dealing with springs. It represents how stiff a spring is and is usually denoted by \( k \). A higher spring constant means a stiffer spring that requires more force to compress or stretch it by a certain amount.

The unit of the spring constant in the international system is Newtons per meter (N/m). In our case, the spring constant is initially given as 35 N/cm. To convert it to the standard unit, we change it to 3500 N/m.

Understanding the spring constant helps us assess how much force a spring can apply, which ties into calculating the potential energy stored in the spring before any compression or expansion. Therefore, the spring constant not only helps us determine stored energy but also predicts the behavior of the spring under different forces.