Question

A 1.35-kg object is attached to a horizontal spring of force constant 2.5 N/cm. The object is started oscillating by pulling it 6.0 cm from its equilibrium position and releasing it so that it is free to oscillate on a frictionless horizontal air track. You observe that after eight cycles its maximum displacement from equilibrium is only 3.5 cm. (a) How much energy has this system lost to damping during these eight cycles? (b) Where did the "lost" energy go? Explain physically how the system could have lost energy.

Short answer

The system lost 0.297 J of energy due to damping, likely dissipated as heat due to air resistance or internal friction.

Step-by-step solution

Convert Units

First, convert the spring constant from N/cm to N/m. Given \( k = 2.5 \) N/cm, to convert to N/m, multiply by 100: \( k = 2.5 \times 100 = 250 \) N/m.

Calculate Initial Potential Energy

Use the formula for the potential energy stored in a spring: \( E_{initial} = \frac{1}{2} k x_0^2 \). Here, \( x_0 = 6.0 \) cm = 0.06 m. Thus, \( E_{initial} = \frac{1}{2} \times 250 \times (0.06)^2 \). Solve this to find the initial energy.

Solve for Initial Energy

Compute the initial potential energy: \( E_{initial} = \frac{1}{2} \times 250 \times 0.0036 = 0.45 \) J.

Calculate Final Potential Energy

The final potential energy after 8 cycles is calculated using \( x = 3.5 \) cm = 0.035 m, \( E_{final} = \frac{1}{2} k x^2 \). This becomes \( E_{final} = \frac{1}{2} \times 250 \times (0.035)^2 \). Solve this to find the final energy.

Solve for Final Energy

Compute the final potential energy: \( E_{final} = \frac{1}{2} \times 250 \times 0.001225 = 0.153125 \) J.

Determine Energy Loss

Subtract the final energy from the initial energy to find the energy lost: \( E_{lost} = E_{initial} - E_{final} = 0.45 - 0.153125 \). Solve this to find the lost energy.

Solve for Energy Loss

Calculate the energy lost: \( E_{lost} = 0.45 - 0.153125 = 0.296875 \) J. Round this to three significant figures for a final answer: \( E_{lost} = 0.297 \) J.

Explain Energy Loss

The energy was lost due to damping forces. Even though the track is frictionless, oscillations can be damped by air resistance or internal friction within the spring and object. The energy was dissipated as heat.

Key concepts

Energy Conservation

Energy conservation is a fundamental concept in physics, stating that energy cannot be created or destroyed, only transformed from one form to another. In the context of damped oscillations, even though we begin with a certain amount of total mechanical energy, some of it may transform into other forms due to damping effects.

In an idealized, frictionless system, like the oscillating spring problem described, one might initially expect no loss of energy. However, in real-world scenarios, energy is almost always exchanged, often with the environment,
leading to a reduction in the mechanical energy of the system.

In our exercise, despite the frictionless track, the energy loss occurs, likely through air resistance or internal friction. These are considered non-conservative forces as they do not store energy but convert it to thermal energy, perceived as a loss.

Potential Energy

Potential energy in a spring-coupled system is stored energy resulting from an object's position relative to equilibrium. This is critical in understanding oscillations, as it determines how 'stretched' or 'compressed' the spring is.

To calculate the potential energy stored in the spring, we use the formula:

• \( E = \frac{1}{2} k x^2 \)...

where \( k \) is the spring constant (in N/m), and \( x \) is the displacement from the equilibrium position (in meters).

In the given exercise, the potential energy reduces over time due to damping. Initially, when the spring is at maximum stretch at 6.0 cm, it has the highest potential energy. After several cycles, with a reduced stretch of 3.5 cm, the potential energy also decreases accordingly. This change in energy helps us understand how much energy the system has dissipated.

Spring Constant

The spring constant \( k \) is a measure of a spring's stiffness. Essentially, it tells us how much force is needed to displace the spring by a certain distance. This parameter is vital as it directly affects the potential energy of the system.

In the exercise, the spring constant is given as 2.5 N/cm, which can be converted to 250 N/m for standard unit calculation. Knowing the spring constant allows us to properly apply the potential energy formula and assess how energy changes as the system oscillates.

The role of the spring constant is significant in determining how a spring will behave under force:

• Higher spring constant: Stiffer spring, harder to stretch, more potential energy when displaced.
• Lower spring constant: Less stiff spring, easier to stretch, less potential energy when displaced.

The constant is crucial for calculating accurate energy values, particularly in exercises where damping plays a role, as it helps quantify how energy is initially stored and subsequently dissipated.