Question

A spring with a spring constant of \(238.5 \mathrm{~N} / \mathrm{m}\) is compressed by \(0.231 \mathrm{~m}\). Then a steel ball bearing of mass \(0.0413 \mathrm{~kg}\) is put against the end of the spring, and the spring is released. What is the speed of the ball bearing right after it loses contact with the spring? (The ball bearing will come off the spring exactly as the spring returns to its equilibrium position. Assume that the mass of the spring can be neglected.)

Short answer

Answer: The speed of the ball bearing right after it loses contact with the spring is approximately 17.54 m/s.

Step-by-step solution

Calculate the elastic potential energy of the compressed spring

According to Hooke's law, the elastic potential energy (PE) stored in a compressed spring is given by the formula: PE = \(\frac{1}{2} kx^2\) where k is the spring constant and x is the displacement from the equilibrium position. We are given k = 238.5 N/m and x = 0.231 m. Let's plug in these values to get the potential energy: PE = \(\frac{1}{2} (238.5 \mathrm{~N/m}) (0.231 \mathrm{~m})^2\)

Compute the potential energy

Now calculate the energy stored in the spring: PE = \(\frac{1}{2} (238.5 \mathrm{~N/m}) (0.053161 \mathrm{~m^2})\) PE = 6.34944825 J

Apply conservation of energy

When the spring is released, the elastic potential energy stored in the spring will be converted to the kinetic energy (KE) of the ball bearing. According to the conservation of energy principle: PE = KE So, 6.34944825 J = \(\frac{1}{2} mv^2\) where m is the mass of the ball bearing (0.0413 kg) and v is its speed.

Solve for the speed of the ball bearing

Now, we need to solve for the speed (v) of the ball bearing: 6.34944825 J = \(\frac{1}{2} (0.0413 \mathrm{~kg}) v^2\) To find the value of v, we can first multiply both sides of the equation by 2, and then divide by the mass (0.0413 kg): v^2 = \(\frac{2 \times 6.34944825 \mathrm{~J}}{ 0.0413 \mathrm{~kg}}\) v^2 = 307.5936211 Now, take the square root of both sides to find the value of v: \(v = \sqrt{307.5936211}\) v ≈ 17.54 m/s

Write the final answer

The speed of the ball bearing right after it loses contact with the spring is approximately 17.54 m/s.

Key concepts

Understanding Hooke's Law

Hooke's Law is a principle of physics that relates the force needed to extend or compress a spring to the distance it is stretched or compressed. It's stated as:
\[ F = -kx \
where \(F\) is the force applied to the spring, \(k\) is the spring constant, and \(x\) is the displacement of the spring from its equilibrium position. The minus sign indicates that the force exerted by the spring is in the opposite direction of the displacement.

To calculate the elastic potential energy stored in a spring, we use the formula:
\[ PE = \frac{1}{2} kx^2 \
which is derived from Hooke's Law. It's essential for students to understand that in physics, 'spring constant' and 'displacement' are not just numbers, but indicate how stiff the spring is and by how much it has been compressed or stretched, respectively. The greater the spring constant, the 'stiffer' or 'stronger' the spring is, and the more force it requires to compress or extend by a given distance. In the provided exercise, the spring constant, \(k = 238.5 \text{N/m}\), and the displacement, \(x = 0.231 \text{m}\), help us to find the elastic potential energy, which can then be used to determine other quantities, like kinetic energy or speed of a moving object.

Conservation of Energy Principle

The Conservation of Energy is a fundamental concept in physics stating that energy cannot be created or destroyed; it can only be transferred or transformed from one form to another. When applying this principle to our problem, the elastic potential energy stored in the spring when it is compressed is converted into kinetic energy as it releases and pushes the ball bearing out.

Mathematically, this is expressed as:
\[ PE_{initial} + KE_{initial} = PE_{final} + KE_{final} \
Since the ball bearing is stationary initially, \(KE_{initial} = 0\). Once the spring returns to its equilibrium position without any further stretching or compression, \(PE_{final} = 0\). Therefore, all of the stored elastic potential energy is converted into kinetic energy (given that no energy is lost to friction or air resistance).

Understanding the conservation of energy allows students to predict the outcome of various physical situations, such as the maximum height a thrown ball will reach, or in our case, the speed of a ball bearing shot from a spring. It's a crucial concept for solving many physics problems and appreciating the consistent behavior of energy in our universe.

Kinetic Energy and Its Calculation

Kinetic Energy (KE) is the energy that an object possesses due to its motion. It is given by the formula:
\[ KE = \frac{1}{2} mv^2 \
where \(m\) is the mass of the object, and \(v\) is its velocity. For the ball bearing in our exercise, once the spring is released and it's in motion, it will have kinetic energy, which we can calculate using the conservation of energy principle.

As the spring releases, the ball bearing speeds up and gains kinetic energy at the expense of the spring's elastic potential energy. By equating the potential energy of the compressed spring to the kinetic energy of the moving ball bearing, we are able to solve for the ball's speed. Here, the mass of the ball bearing (\(m = 0.0413 \text{kg}\)) is an essential factor in calculating the kinetic energy.

Understanding kinetic energy is important for students because it applies to everything in motion, from microscopic particles to astronomical objects, and it allows learners to solve practical problems involving the motion of objects. The final calculation ensures students can connect the stored energy in an object at rest with the dynamic energy of the same object in motion.