Question:

A 2 kg object is initially at a height of 5 m above the ground. It is then released and falls downward. Assume there is no air resistance. What is the object's final velocity just before it hits the ground?

Answer:

We can solve this problem using the principle of conservation of energy, which states that the total energy of a closed system is constant. In this case, the system consists of the object and the Earth.

We can consider the system at two points: at the initial height and just before the object hits the ground.

At the initial height, the object has gravitational potential energy given by:

$PE_{initial} = m \cdot g \cdot h_{initial}$

Where:

• $PE_{initial}$ is the gravitational potential energy at the initial height,
• $m$ is the mass of the object (2 kg),
• $g$ is the acceleration due to gravity (9.8 m/s²),
• $h_{initial}$ is the initial height (5 m).

Substituting the values:

$PE_{initial} = 2 \, \text{kg} \cdot 9.8 \, \text{m/s²} \cdot 5 \, \text{m}$

$PE_{initial} = 98 \, \text{J}$

At the final point just before the object hits the ground, all of the object's initial gravitational potential energy is converted into kinetic energy:

$PE_{final} = KE_{final}$

$m \cdot g \cdot h_{final} = \frac{1}{2} \cdot m \cdot v_{final}^2$

Where:

• $PE_{final}$ is the gravitational potential energy at the final point,
• $KE_{final}$ is the kinetic energy at the final point,
• $h_{final}$ is the final height (0 m, as the object hits the ground),
• $v_{final}$ is the final velocity of the object.

Simplifying the equation:

$g \cdot h_{final} = \frac{1}{2} \cdot v_{final}^2$

$9.8 \, \text{m/s²} \cdot 0 \, \text{m} = \frac{1}{2} \cdot v_{final}^2$

Since $0 \cdot a = 0$ for any value of $a$, we can conclude that the final velocity of the object just before hitting the ground is 0 m/s.

Therefore, the object's final velocity just before it hits the ground is 0 m/s.