Question

A \(5.76-\mathrm{kg}\) rock is dropped and allowed to fall freely. Find the initial kinetic energy, the final kinetic energy, and the change in kinetic energy for (a) the first \(2.00 \mathrm{~m}\) of fall and (b) the second \(2.00 \mathrm{~m}\) of fall.

Short answer

For the first 2.00 m: Initial KE = 0 J, Final KE = 112.8 J, Change = 112.8 J. For the second 2.00 m: Initial KE = 112.8 J, Final KE = 224.0 J, Change = 111.2 J.

Step-by-step solution

Understand the Problem

We need to find the kinetic energy of a \(5.76 \mathrm{~kg}\) rock during its fall. The problem is divided into two scenarios: the first \(2.00 \mathrm{~m}\) of fall and the second \(2.00 \mathrm{~m}\) of fall. We will calculate the initial kinetic energy, the final kinetic energy, and the change in kinetic energy for both cases. We will use the formula for gravitational potential energy conversion to kinetic energy to find these values.

Calculate Initial and Final Velocity for the First 2.00 m

For the first \(2.00 \mathrm{~m}\) of fall from rest, the initial velocity \(v_0\) is \[v_0 = 0 \mathrm{~m/s}.\]Use the equation of motion \(v^2 = u^2 + 2as\) to find the velocity \(v\) after \(2.00 \mathrm{~m}\), where \(a = 9.81 \mathrm{~m/s^2}\) is the acceleration due to gravity and \(s = 2.00 \mathrm{~m}\) is the distance:\[v^2 = 0 + 2 \times 9.81 \times 2.00.\]Solving gives:\[v = \sqrt{2 \times 9.81 \times 2.00} = 6.26 \mathrm{~m/s}.\]

Calculate Kinetic Energies for the First 2.00 m

The initial kinetic energy \(KE_i\) is \[KE_i = \frac{1}{2} m v_0^2 = \frac{1}{2} \times 5.76 \times 0^2 = 0 \mathrm{~J}. \]The final kinetic energy \(KE_f\) after falling \(2.00 \mathrm{~m}\) is \[KE_f = \frac{1}{2} m v^2 = \frac{1}{2} \times 5.76 \times (6.26)^2 = 112.8 \mathrm{~J}. \]The change in kinetic energy is \[\Delta KE = KE_f - KE_i = 112.8 - 0 = 112.8 \mathrm{~J}.\]

Calculate Initial and Final Velocity for the Second 2.00 m

The initial velocity at the start of the second \(2.00 \mathrm{~m}\) fall is \(6.26 \mathrm{~m/s}\) (from the final velocity of the first \(2.00 \mathrm{~m}\) fall).Use \(v^2 = u^2 + 2as\) again to find the new velocity after an additional \(2.00 \mathrm{~m}\):\[v^2 = 6.26^2 + 2 \times 9.81 \times 2.00.\]Solving gives:\[v = \sqrt{6.26^2 + 2 \times 9.81 \times 2.00} = 8.84 \mathrm{~m/s}.\]

Calculate Kinetic Energies for the Second 2.00 m

The initial kinetic energy \(KE_i\) is \[KE_i = \frac{1}{2} m (6.26)^2 = 112.8 \mathrm{~J}.\]The final kinetic energy \(KE_f\) after the second \(2.00 \mathrm{~m}\) is \[KE_f = \frac{1}{2} m v^2 = \frac{1}{2} \times 5.76 \times (8.84)^2 = 224.0 \mathrm{~J}.\]The change in kinetic energy for this segment is \[\Delta KE = KE_f - KE_i = 224.0 - 112.8 = 111.2 \mathrm{~J}.\]

Key concepts

Free Fall

When we talk about free fall, we're talking about an object moving under the influence of gravity alone. No other forces, like air resistance, are taken into account. Imagine dropping a rock from a height. The moment you let go, gravity takes over, pulling it towards the Earth. This is a classic example of free fall.

Gravity essentially acts as a natural accelerator, giving everything a downward acceleration of approximately \[9.81 \ \text{m/s}^2.\] This means that, as the rock falls, it speeds up by about 9.81 meters per second every second.

In the exercise, the rock starts with no initial speed. As it begins its journey downwards, its speed increases, demonstrating how gravity relentlessly fosters motion in free-falling objects. By considering just gravity, calculations in free fall become clearer and help us focus on how weight and height influence movement.

Velocity Calculation

Velocity calculation in free-fall scenarios involves understanding how quickly an object is moving at a given point during its descent. We use basic physics principles to determine velocity at specific distances.

For our rock example, we initially have \[v_0 = 0 \ \text{m/s}.\] As it falls, we want to know the velocity at certain intervals, like after falling the first 2.00 meters. To calculate this, the equation we use is:\[v^2 = u^2 + 2as,\]where

• \(u\) is the initial velocity,
• \(a\) is the acceleration due to gravity (9.81 m/s²), and
• \(s\) is the distance fallen.

Evaluating this for the first 2.00 meters gives us a new velocity of roughly 6.26 m/s.

For the next 2.00 meters, the cycle repeats. The new initial velocity is the final velocity of the previous phase, and it increases further due to gravity. This calculation is crucial for determining kinetic energy changes, as kinetic energy directly depends on velocity.

Gravitational Potential Energy

Gravitational potential energy (GPE) is all about elevation and mass. It's the energy an object possesses because of its position relative to Earth. In any free fall situation, like our rock, GPE at the start is converted into kinetic energy (KE) as it falls.

The formula for GPE is:\[PE = mgh,\]where

• \(m\) is the mass of the object,
• \(g\) is the acceleration due to gravity, and
• \(h\) is the height relative to a reference point.

At the highest point, the rock has maximum potential energy and zero kinetic energy since it's not moving yet. As it falls, the height decreases, leading to a conversion of GPE into KE, increasing the rock's speed.

Understanding this exchange between gravitational potential energy and kinetic energy helps not only in calculating the speed but also in comprehending how energy balance is maintained during the fall. This transformation is central to many physical phenomena and real-world applications.