Understanding the Physics of a Dropped Rigid Body

This problem involves analyzing the motion and energy of a rigid body falling under the influence of gravity. We need to determine its speed just as it reaches the ground and its total mechanical energy when it is dropped from a specific height. The key principles involved are kinematics under constant acceleration (due to gravity) and the conservation of mechanical energy.

Given Information:

• Mass of the rigid body, \(m = 2 \, \text{kg}\)
• Initial height from the ground, \(h = 50 \, \text{m}\)
• Initial velocity, \(u = 0 \, \text{m/s}\) (since it's dropped from rest)
• Acceleration due to gravity, \(g = 9·8 \, \text{m/s}^2\)

Calculating the Speed When the Body Just Touches the Ground

We can find the final velocity (\(v\)) just before hitting the ground using kinematic equations or the principle of conservation of mechanical energy.

Method 1: Using Kinematic Equation

The relevant kinematic equation relating initial velocity (\(u\)), final velocity (\(v\)), acceleration (\(a\)), and displacement (\(s\)) is:

\(v^2 = u^2 + 2as\)

In this case:

• \(u = 0 \, \text{m/s}\)
• \(a = g = 9·8 \, \text{m/s}^2\) (positive as direction is downwards)
• \(s = h = 50 \, \text{m}\)

Substituting the values:

\(v^2 = (0)^2 + 2 \times (9·8 \, \text{m/s}^2) \times (50 \, \text{m})\)

\(v^2 = 0 + 2 \times 490 \, \text{m}^2/\text{s}^2\)

\(v^2 = 980 \, \text{m}^2/\text{s}^2\)

Taking the square root to find \(v\):

\(v = \sqrt{980} \, \text{m/s}\)

Method 2: Using Conservation of Mechanical Energy

Mechanical energy is conserved if only conservative forces (like gravity) are doing work. The total mechanical energy (\(E\)) is the sum of kinetic energy (KE) and potential energy (PE).

\(E = KE + PE\)

Initial mechanical energy (\(E_{initial}\)) when dropped from height \(h\):

\(KE_{initial} = \frac{1}{2}mu^2 = \frac{1}{2}(2 \, \text{kg})(0 \, \text{m/s})^2 = 0 \, \text{J}\)

\(PE_{initial} = mgh = (2 \, \text{kg})(9·8 \, \text{m/s}^2)(50 \, \text{m}) = 980 \, \text{J}\)

\(E_{initial} = KE_{initial} + PE_{initial} = 0 \, \text{J} + 980 \, \text{J} = 980 \, \text{J}\)

Final mechanical energy (\(E_{final}\)) just before hitting the ground (height \(h = 0\)):

\(KE_{final} = \frac{1}{2}mv^2 = \frac{1}{2}(2 \, \text{kg})v^2 = v^2 \, \text{J}\)

\(PE_{final} = mgh = (2 \, \text{kg})(9·8 \, \text{m/s}^2)(0 \, \text{m}) = 0 \, \text{J}\)

\(E_{final} = KE_{final} + PE_{final} = v^2 \, \text{J} + 0 \, \text{J} = v^2 \, \text{J}\)

By conservation of mechanical energy, \(E_{initial} = E_{final}\):

\(980 \, \text{J} = v^2 \, \text{J}\)

\(v^2 = 980 \, \text{m}^2/\text{s}^2\)

\(v = \sqrt{980} \, \text{m/s}\)

Both methods give the same speed at the ground: \(\sqrt{980} \, \text{m/s}\).

Calculating the Total Energy When Dropped

As calculated in Method 2 for the speed, the total energy when the body is dropped from the balloon is the sum of its kinetic and potential energy at that height.

Initial kinetic energy (\(KE_{initial}\)) is 0 as the body starts from rest.

\(KE_{initial} = 0 \, \text{J}\)

Initial potential energy (\(PE_{initial}\)) at height \(h=50 \, \text{m}\) is:

\(PE_{initial} = mgh = (2 \, \text{kg})(9·8 \, \text{m/s}^2)(50 \, \text{m}) = 980 \, \text{J}\)

Total energy when dropped (\(E_{initial}\)) is:

\(E_{initial} = KE_{initial} + PE_{initial} = 0 \, \text{J} + 980 \, \text{J} = 980 \, \text{J}\)

The total energy when dropped is 980 J.

Summary of Results

• Speed when it just touches the ground: \(\sqrt{980} \, \text{m/s}\)
• Total energy when it is dropped: 980 J

Therefore, the speed and total energy are respectively \(\sqrt{980} \, \text{m/s}\) and 980 J.

Revision Table: Key Concepts

Additional Information: Energy Transformation During Free Fall

As the rigid body falls from the balloon, its energy transforms. Initially, at the maximum height, its speed is zero, so its energy is purely potential energy. As it falls, the height decreases, reducing potential energy, while its speed increases, increasing kinetic energy. This transformation continues until just before hitting the ground, where the height is zero, and the potential energy is zero. At this point, all the initial potential energy has been converted into kinetic energy (assuming no energy loss due to air resistance).

The total mechanical energy (\(KE + PE\)) remains constant throughout the fall, demonstrating the principle of conservation of energy in a closed system under conservative forces.

In this problem, the initial total energy was 980 J (all potential), and the final total energy just before impact is also 980 J (all kinetic).