Mechanical Energy and Work

Introduction to Work

In physics, work is done when a force causes displacement. It is a measure of energy transfer that occurs when an object is moved by a force.

Work is only done when the object actually moves in the direction of the applied force. If there is no movement, or if the force is perpendicular to the motion, no work is done.

The general formula for work using vectors is:

\[ W = \vec{F} \cdot \vec{d} = Fd \cos \theta \]

\( \vec{F} \) = applied force vector
\( \vec{d} \) = displacement vector
\( \theta \) = angle between force and displacement

This formula is known as the dot product (scalar product), which calculates how much of the force acts in the direction of motion.

Key Conditions for Work to Be Done

A force must be applied.
The object must move.
The force must have a component along the direction of motion (if \( \theta = 90^\circ \), no work is done).

Types of Work

Positive Work (W > 0): When the force and displacement are in the same direction (e.g., lifting an object).
Negative Work (W < 0): When the force opposes the displacement (e.g., friction slowing down an object).
Zero Work (W = 0): When there is no movement or force is perpendicular to displacement.

Energy is the ability to do work. It is a fundamental quantity in physics that can exist in various forms and can be transferred or transformed but not created or destroyed (according to the law of conservation of energy).

You are already familiar with Energy, Conservation of Energy, Forms of Energy and Sources of Energy from the previous chapters.

Types of Mechanical Energy

Kinetic Energy (KE): \( KE = \frac{1}{2}mv^2 \)
Potential Energy (PE):
- Gravitational PE: \( PE = mgh \)
- Elastic PE: \( PE = \frac{1}{2}kx^2 \)

Relationship Between Work Done and Energy

The relationship between them is described by the work-energy theorem, which states:

\( \text{Work done on an object} = \text{Change in its energy} \)

\( W = \Delta E \)

A person pushes a 10 kg box with a force of 20 N over a distance of 5 m on a frictionless surface. How much work is done by the person?

\( W = Fd \cos \theta = 20 \times 5 \times \cos 0 = 100 \, \text{J} \)

A ball of mass 2 kg is thrown vertically upward with an initial velocity of 15 m/s. How high will it rise before stopping?

Using conservation of energy:

\( KE_i = GPE_f \)

\( \frac{1}{2}mv^2 = mgh \Rightarrow \frac{1}{2}(2)(15^2) = 2 \cdot 9.8 \cdot h \)

\( h = 11.48 \, \text{m} \)

A 50 kg cyclist is moving at 10 m/s. If the cyclist stops by applying brakes that exert a constant friction force of 200 N, what is the stopping distance?

\( \Delta KE = \text{Work Done} = Fd \Rightarrow \frac{1}{2}mv^2 = Fd \)

\( \frac{1}{2} \cdot 50 \cdot 10^2 = 200d \Rightarrow d = 12.5 \, \text{m} \)

A pendulum is released from rest at a height of 2 m. What will be its speed at the lowest point?

\( GPE = KE \Rightarrow mgh = \frac{1}{2}mv^2 \Rightarrow v = \sqrt{2gh} = \sqrt{2 \cdot 9.8 \cdot 2} = 6.26 \, \text{m/s} \)

Written by Thenura Dilruk