91Ó°ÊÓ

The concepts of work and energy are essential in physics, particularly in mechanics, where they help us to understand how forces cause displacement and changes in velocity. Work, in a physical sense, is defined as the force applied over a certain displacement. The basic formula for work is given by:

• \( W = F \cdot s \cdot \cos(\theta) \)

in which \( F \) represents the force, \( s \) denotes the displacement, and \( \theta \) is the angle between the force and the displacement direction. When the direction of the force aligns with the displacement, the angle \( \theta \) becomes zero, and thus \( \cos(\theta) = 1 \).
Energy is the capacity to do work. The principle of conservation of energy states that energy cannot be created or destroyed, but only transformed from one form to another. In the context of the helicopter problem, we're looking at mechanical energy, which includes kinetic and potential energy. When the helicopter ascends, work is performed by the lifting force, increasing the helicopter's potential energy due to its elevation against gravity.

Newton's laws of motion form the cornerstone of classical mechanics, describing the relationship between a body and the forces acting on it. In this particular problem with the helicopter, Newton's Second Law is of primary focus. It states that:

• \( F = ma \)

Here, \( F \) is the net force acting on the body, \( m \) is the mass, and \( a \) is the acceleration.
For the helicopter ascending with a specific acceleration, the lifting force must first overcome the gravitational force acting downwards, known as weight, which is calculated by \( mg \), where \( g \) is the acceleration due to gravity (approximately \(9.81 \text{ m/s}^2\)). Any additional force will contribute to the upward acceleration. Thus, the resultant lifting force on the helicopter is the sum of the forces necessary to overcome gravity and to provide the additional acceleration.

Determining displacement under constant acceleration is crucial in motion-related physics problems. When the helicopter rises, it does so with a uniform acceleration. We utilize the formula:

• \( s = ut + \frac{1}{2}at^2 \)

where \( s \) is the displacement, \( u \) is the initial velocity, \( a \) is the acceleration, and \( t \) represents time.
In this problem, the helicopter starts from rest, meaning \( u = 0 \) m/s. The equation simplifies to \( s = \frac{1}{2}at^2 \). This allows us to calculate that the helicopter travels 25 meters upwards in 5 seconds, given the specified acceleration. This displacement is used to determine the work done by forces on the helicopter.

Gravitational force is the force of attraction between two masses. On Earth, it gives weight to physical objects, pulling them toward the center. The force is given by the formula:

• \( F_g = mg \)

where \( m \) is the mass of the object and \( g \) is the acceleration due to gravity, approximately \( 9.81 \text{ m/s}^2 \) at Earth's surface.
For the helicopter, the gravitational force is calculated to be 4905 N and acts downward. When calculating work done by gravity as the helicopter ascends, the force opposes the direction of movement. Hence, the work done by gravity is negative, as determined by the expression \( W = F_g \cdot s \cdot \cos(\theta) \), with \( \theta = 180° \), which brings the cosine term to \(-1\). This negative work signifies that gravitational force is doing work against the helicopter's motion, decreasing its net mechanical work.