Calculating Work Done by Gravity on a Falling Ball

This question asks us to determine the work done by the force of gravity on a ball as it falls through a specific distance. Understanding the concept of work done by a force is key here.

What is Work Done?

In physics, work done by a constant force is defined as the product of the magnitude of the force, the magnitude of the displacement, and the cosine of the angle between the force and the displacement vectors. Mathematically, it is given by:

\(W = F \cdot d \cdot \cos(\theta)\)

• \(W\) is the work done.
• \(F\) is the magnitude of the force.
• \(d\) is the magnitude of the displacement.
• \(\theta\) is the angle between the force vector and the displacement vector.

Applying the Concepts to the Problem

We are given the following information about the ball and its motion:

• Mass of the ball (\(m\)) = 0.1 kg
• Distance fallen (\(d\)) = 2 m
• Acceleration due to gravity (\(g\)) = 9.8 m/s\(^2\)

The force acting on the ball in this scenario is the force of gravity, which is the weight of the ball. The weight (\(F_g\)) is calculated as:

\(F_g = m \cdot g\)

Let's calculate the magnitude of the force of gravity:

\(F_g = 0.1 \text{ kg} \times 9.8 \text{ m/s}^2 = 0.98 \text{ N}\)

The ball is dropped and falls downwards. The force of gravity also acts downwards. Therefore, the direction of the force of gravity and the direction of the displacement are the same.

The angle (\(\theta\)) between the force of gravity vector (downwards) and the displacement vector (downwards) is 0 degrees.

Now we can calculate the work done by the force of gravity using the work formula:

\(W = F_g \cdot d \cdot \cos(\theta)\)

\(W = 0.98 \text{ N} \times 2 \text{ m} \times \cos(0^{\circ})\)

Since \(\cos(0^{\circ}) = 1\), the equation simplifies to:

\(W = 0.98 \text{ N} \times 2 \text{ m} \times 1\)

\(W = 1.96 \text{ J}\)

The work done by the force of gravity is positive because the force and the displacement are in the same direction. Gravity is doing positive work, transferring energy to the ball (increasing its kinetic energy as it falls).

Summary of Calculation:

The calculated work done by the force of gravity is 1.96 J.

Conclusion

Based on the calculation, the work done by the force of gravity when the 0.1 kg ball falls through a distance of 2 m is 1.96 J.

Revision Table: Work Done by Forces

Additional Information: Work-Energy Theorem and Potential Energy

The concept of work done by gravity is closely related to potential energy and the work-energy theorem.

• Gravitational Potential Energy: For an object near the Earth's surface, the change in gravitational potential energy (\(\Delta PE_g\)) when it falls a distance \(h\) is given by \(\Delta PE_g = -mgh\). The work done by gravity is equal to the negative of the change in potential energy, i.e., \(W_g = -\Delta PE_g\). In this case, as the ball falls, its potential energy decreases (\(\Delta PE_g\) is negative), so the work done by gravity is positive (\(W_g = -(-1.96 \text{ J}) = 1.96 \text{ J}\)).
• Work-Energy Theorem: This theorem states that the net work done on an object by all forces is equal to the change in its kinetic energy (\(\Delta KE\)). If gravity is the only significant force doing work on the falling ball (ignoring air resistance), then the work done by gravity equals the change in the ball's kinetic energy. As the ball falls from rest, its kinetic energy increases from zero.

These concepts reinforce that gravity does positive work on a falling object, transferring energy and increasing its speed.