Calculating Velocity Using Conservation of Energy: An Example with a Falling Body

When dealing with problems involving falling bodies, one of the most commonly used principles in physics is the conservation of energy. This principle states that the total energy in a system remains constant, assuming no external work is done on the system. In this article, we will explore how to calculate the velocity of a body using the conservation of energy approach, as demonstrated in a problem where a 2 kg body is dropped from a height of 5 meters above the ground.

Problem Statement

A body of mass 2 kg is dropped from a rest position 5 meters above the ground. What is its velocity at a height of 3 meters above the ground?

Preliminary Concepts: Energy and Energy Conversion

Before diving into the calculation, itrsquo;s essential to understand the concepts of potential energy and kinetic energy.

Potential Energy (PE)

Potential energy is the energy an object possesses due to its position or configuration. For a body near the Earthrsquo;s surface, the gravitational potential energy (PE) is given by the formula:

[ PE mgh ]

where:

(m) is the mass of the object, (g) is the acceleration due to gravity (approximately 9.81 m/s2), (h) is the height above the reference point.

Kinetic Energy (KE)

Kinetic energy is the energy 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, (v) is the velocity of the object.

Step-by-Step Calculation

Given data:

Mass of the body, (m 2, text{kg}) Initial height, (h_i 5, text{m}) Final height, (h_f 3, text{m}) Initial velocity, (v_i 0, text{m/s}) (since the body is dropped from rest)

Calculate the initial potential energy (PEinitial):

[ PE_i mgh_i 2, text{kg} times 9.81, text{m/s}^2 times 5, text{m} 98.1, text{J} ]

Calculate the final potential energy (PEfinal):

[ PE_f mgh_f 2, text{kg} times 9.81, text{m/s}^2 times 3, text{m} 58.86, text{J} ]

Calculate the change in potential energy (ΔPE):

[ Delta PE PE_i - PE_f 98.1, text{J} - 58.86, text{J} 39.24, text{J} ]

Since the change in potential energy is converted to kinetic energy (KE), we can set

[ Delta PE KE ]

Solve for the final velocity (v):

[ KE frac{1}{2}mv^2 ]

Substitute the change in potential energy:

[ 39.24, text{J} frac{1}{2} times 2, text{kg} times v^2 ]

Simplify and solve for (v):

[ 39.24 v^2 ]

[ v sqrt{39.24} approx 6.26, text{m/s} ]

Conclusion

The velocity of the body at a height of 3 meters above the ground is approximately 6.26 m/s.

Welcome to the world of physics, where understanding the principles like conservation of energy can help solve complex real-world problems. By applying the correct formulas and principles, we can determine the kinetic energy and velocity of a falling object with ease.