Work, Energy, and Simple Machines: Work Done by a Constant Force

Introduction to Work, Energy, and Power

• Limitations of Kinematics and Newton's Laws:

  • Kinematic equations and Newton's laws of motion provide tools to analyze how forces change the motion of objects.

  • When forces change with time or act in complicated, multidirectional ways, applying Newton's laws directly becomes computationally difficult.

  • Work, energy, and power offer a simpler and more powerful framework for analyzing complex physical motion and interactions.

• Role of Energy and Simple Machines:

  • Energy is defined as the capacity to do work and lies at the foundation of all physical activities.

  • Common tasks require energy from varied physical sources:

  • Biological chemical energy from food provides the energy necessary to walk.

  • Electrical energy provides the energy needed to rotate a ceiling fan.

  • Chemical potential energy in fuel provides the mechanical energy required to move an automobile.

  • Simple machines serve as essential building blocks that enable tasks to be performed with less effort and greater convenience.

Concept of Work Done by a Constant Force

• Lifting Objects and Physical Observations:

  • Consider a wheat bag of mass $m = 5\,\text{kg}$ placed flat on the floor.

  • A downward gravitational force equal to $mg$ acts on the bag, where $m$ is the mass and $g$ is the acceleration due to gravity.

  • Lifting the bag slowly to a height of $s = 1\,\text{m}$ requires applying an upward force $F$ equal in magnitude to its weight ($F = mg$).

  • The applied force acts upward as the bag undergoes an upward displacement of $1\,\text{m}$ in the direction of the force.

• Proportionality Principles of Work:

  • Lifting Multiple Objects Sequentially:

  • Lifting three identical $5\,\text{kg}$ bags one after another to a height of $1\,\text{m}$ requires doing three times as much work ($3W$) as lifting one bag.

  • If a fuel-powered machine carries out this task, it consumes three times more fuel to lift three bags sequentially.

  • Lifting Multiple Objects Simultaneously:

  • Lifting all three $5\,\text{kg}$ bags together to a height of $1\,\text{m}$ requires applying an upward force three times larger ($3F = 3mg$).

  • Because the net physical task matches sequential lifting, the total work done equals $3W$.

  • Principle: Applying a larger force over the same distance yields a directly proportional increase in work done.

  • Lifting over Greater Distances:

  • Lifting a single $5\,\text{kg}$ bag to a vertical height of $s = 3\,\text{m}$ requires three times more work ($3W$) than lifting it by $1\,\text{m}$.

  • Operating a lifting machine three successive times to raise the bag by $1\,\text{m}$ each time consumes three times more fuel.

  • Principle: Applying the same force over a larger distance yields a directly proportional increase in work done.

Mathematical Formulation and Units of Work

• Scientific Definition of Work:

  • The work done on an object by a constant force is defined as the product of the force applied and the displacement achieved in the direction of the force:     $\text{Work done on an object by a constant force} = \text{force applied} \times \text{displacement in the direction of the force}$

  • When a constant force $F$ acts on an object and causes a displacement $s$ in the direction of the force, the mathematical equation for work done $W$ is:     $W = F \times s$

  • Equation $W = F \times s$ applies universally whether displacement occurs in vertical, horizontal, or inclined directions, provided displacement is along the line of action of the force.

  • Specification Requirement: When describing work done, it is essential to explicitly specify both the force (or agency) performing the work and the specific object on which the work is done.

• SI Units and Dimensional Analysis:

  • The SI unit of work done is the joule, represented by the symbol $\text{J}$.

  • The SI unit of force is the newton ($\text{N}$), and the SI unit of displacement is the metre ($\text{m}$).

  • Unit conversion relation:     $1\,\text{J} = 1\,\text{N} \times 1\,\text{m}$

  • Definition of 1 Joule: Exactly 1 joule of work is done on an object when a constant force of 1 newton is applied to it and displaces it by 1 metre in the direction of the force.

  • Dimensional derivation using base SI units ($1\,\text{N} = 1\,\text{kg}\,\text{m}\,\text{s}^{-2}$):     $1\,\text{J} = 1\,\text{kg}\,\text{m}\,\text{s}^{-2} \times 1\,\text{m} = 1\,\text{kg}\,\text{m}^2\,\text{s}^{-2}$

Graphical Representation of Work Done

• Determining Work from a Force-Displacement Graph:

  • Work done by a force can be graphically evaluated by plotting Force ($\text{N}$) on the vertical Y-axis against Displacement ($\text{m}$) in the direction of force on the horizontal X-axis.

  • For a constant force, the plot forms a horizontal line, creating a rectangular shape bounded by the force line and displacement interval.

  • The numerical value of work done on the object equals the area under the force-displacement curve:     $\text{Area of shaded region} = 10\,\text{N} \times 1.0\,\text{m} = 10\,\text{J}$

  • When force varies with displacement (non-constant force), work done is calculated by evaluating the total area under the force-displacement curve between the initial and final position markers.

Conditions for Zero Work Done

• Zero Applied Force ($F = 0$):

  • If the force acting on an object is zero ($F = 0$), the work done by that force is strictly zero ($W = 0$).

• Zero Displacement ($s = 0$):

  • If an object undergoes zero displacement ($s = 0$), no work is done on the object regardless of the magnitude of force exerted.

  • Example: Applying an intense pushing force against a rigid, immobile brick wall.

  • Physiological vs Scientific Work: Pushing an immovable wall causes muscular fatigue because internal body muscle fibers repeatedly expand and contract, consuming internal metabolic energy. However, in physics terms, because the wall remains at $s = 0\,\text{m}$, zero work is done on the wall.

• Perpendicular Force and Displacement ($\text{Force} \perp \text{Displacement}$):

  • If an applied force acts perpendicular ($90^\circ$) to the displacement vector of an object, the work done by that force is zero because there is zero displacement along the line of action of the force.

  • Example: A girl carrying a box while walking horizontally across a flat floor.

  • The girl applies an upward force to balance the downward weight of the box.

  • The box displaces horizontally forward.

  • Because the upward force vector and horizontal displacement vector are perpendicular to one another, the carrying force performs zero work on the box.

Positive and Negative Work Done

• Scalar Characteristics of Work:

  • Force and displacement are vector quantities that possess both magnitude and direction.

  • Work is a scalar quantity; it lacks directional orientation, but can take positive, negative, or zero values based on vector alignment.

• Positive Work Done:

  • Occurs when object displacement is in the same direction as the applied force vector.

  • Example: Pushing a wheelchair forward, where the applied force and wheelchair displacement share an identical forward direction.

• Negative Work Done:

  • Occurs when object displacement is in the opposite direction to the applied force vector.

  • Example: A goalkeeper stopping a soccer ball exerts a retarding force opposite to the ball's forward direction of motion and displacement.

• Newton's Third Law and Work Reciprocity:

  • In interacting systems (such as a goalkeeper catching a ball), both entities exert equal and opposite forces on each other under Newton's third law.

  • The moving ball exerts a forward force on the goalkeeper's hands in the direction of backward movement, doing positive work on the goalkeeper.

  • The goalkeeper exerts a retarding force backward against the ball's forward motion, doing negative work on the ball.

Worked Examples and Analytical Questions

• Dumbbell Exercise Analysis (Example 7.1):

  • Scenario: A girl lifts a dumbbell upward during exercise and then slowly lowers it down.

  • Lifting Phase: The girl exerts an upward force matching the dumbbell's weight. Displacement occurs upward. Because force and displacement act in the same direction, work done by the girl on the dumbbell is positive.

  • Lowering Phase: The girl continues applying an upward supporting force to control the dumbbell while displacement occurs downward. Because applied force and displacement act in opposite directions, work done by the girl on the dumbbell is negative.

• Goalkeeper Retarding Work Calculation (Example 7.2):

  • Problem Statement: A goalkeeper stops a soccer ball by applying a retarding force of $200\,\text{N}$, during which her hands recoil backward by $15\,\text{cm}$. Calculate the work done by the goalkeeper on the ball.

  • Step 1: Convert units to standard SI format:     $s = 15\,\text{cm} = 0.15\,\text{m}$

  • Step 2: Assign displacement sign based on vector direction:

  • The ball displaces opposite to the retarding force applied by the goalkeeper, making displacement $-0.15\,\text{m}$ relative to applied force direction.

  • Step 3: Calculate work using $W = F \times s$:     $W = 200\,\text{N} \times (-0.15\,\text{m}) = -30\,\text{J}$

  • Conclusion: The goalkeeper performs $-30\,\text{J}$ of work on the ball to bring it to rest.

• Pause and Ponder Solutions:

  • Question 1: Is a weightlifter doing any work on a barbell while holding it steady above her head?

  • Answer: No. Because the steady barbell undergoes zero displacement ($s = 0\,\text{m}$), the work done on the barbell by the weightlifter is zero ($W = 0\,\text{J}$).

  • Question 2: Is the work done by kinetic friction on a stack of coins sliding across a rough surface positive, negative, or zero?

  • Answer: Negative. Kinetic friction acts in a direction directly opposite to the sliding displacement of the coins. Thus, work done by friction is negative.