Work Done by Constant & Variable Forces - Ninth Grade Science
In everyday language, the word "work" is applied loosely to any activity requiring physical exertion, mental concentration, or commercial employment. You might claim to have worked intensely while reading a dense chemistry textbook, holding a heavy suitcase stationary while waiting for a bus, or pushing against a locked door that refuses to budge. In the precise language of physics, however, none of these activities qualify as work. Physical work requires not merely the expenditure of muscular energy or biological fatigue; it demands that a mechanical force produce an actual spatial displacement along the direction of that force. If there is no displacement, or if the displacement is perpendicular to the applied force, the scientific work done is strictly zero. In this chapter, you will investigate the scientific definition of work, analyze work done under constant and inclined forces, explore negative and zero work, and examine work done by variable forces using graphical integration.
The Scientific Conception of Work
To establish whether work has been accomplished in a physical system, two indispensable conditions must be satisfied simultaneously:
1. A physical force must act upon the object.
2. The object must undergo a spatial displacement, and this displacement must have a non-zero component along the line of action of the applied force.
+--------------------------------------------------------------------------+
| The Two Mandatory Criteria for Work |
+--------------------------------------------------------------------------+
| 1. Force Condition: A net external force F must be applied. |
| 2. Displacement Condition: The object must move through displacement s. |
| If F = 0, Work = 0. If s = 0, Work = 0. |
+--------------------------------------------------------------------------+
Scientific Work vs Everyday "Work"
Case A: Pushing a Rigid Concrete Wall Case B: Pushing a Wheeled Cart
Force = 300 N, Displacement = 0 m Force = 50 N, Displacement = 10 m
+-------+ +-------+
300 N | Wall | 50 N | Cart | ----> 10 m
======> | Rigid | ======> | (o-o) |
+-------+ +-------+
Work Done = 0 Joules! Work = 50 N * 10 m
(Exhausting, but NO physical work) = 500 Joules!
If a student holds a twenty-kilogram barbell motionless above their head for ten minutes, their skeletal muscles burn chemical energy to maintain tension, producing biological exhaustion; yet, because the displacement of the barbell is zero ($s = 0$), the mechanical work performed on the barbell is zero joules.
Mathematical Formulation: Work by a Constant Force
When a constant force $F$ acts on an object and causes it to move through a straight-line displacement $s$ in the exact same direction as the force, the work done ($W$) is defined as the product of the force magnitude and the displacement:
W = F \times s
The SI Unit of Work: The Joule (J)
Because work is the product of force (newtons) and displacement (meters), its standard SI unit is the newton-meter ($ ext{N}\cdot ext{m}$), which is officially named the joule (J) in honor of the English physicist James Prescott Joule:
1\text{ Joule (J)} = 1\text{ Newton} \times 1\text{ Meter} = 1\text{ N}\cdot\text{m} = 1\text{ kg}\cdot\text{m}^2/\text{s}^2
One joule is defined as the amount of work done when a constant force of one newton displaces an object through a distance of one meter in the direction of the force. Larger commercial or geological energy transfers use kilojoules ($1 ext{ kJ} = 10^3 ext{ J}$) and megajoules ($1 ext{ MJ} = 10^6 ext{ J}$).
Scalar Nature of Work
Although work is calculated from two vector quantities—force and displacement—work itself is fundamentally a scalar quantity. It possesses magnitude and an algebraic sign (positive, negative, or zero), but it has no spatial direction in three-dimensional space. You cannot perform "five joules of work to the north."
Work Done by a Force at an Angle: $W = F s \cos heta$
In many practical situations, the applied force does not align directly with the direction of motion. For example, when a child pulls a toy wagon using an inclined handle, the pulling force acts upward and forward at an angle $ heta$, while the wagon rolls horizontally across the ground.
Force Exerted at an Angle theta
Force Vector (F)
/|
/ |
/ | F * sin(theta) [Perpendicular: No Work]
/ |
+----+
F * cos(theta) [Parallel to Motion]
+-----------+
| Toy Wagon | --------------> Displacement (s)
+-----------+
Resolving Force into Vector Components
To determine the work done, we resolve the force vector $F$ into two perpendicular rectangular components:
1. The parallel component, $F_{\parallel} = F \cos heta$, acts parallel to the horizontal displacement.
2. The perpendicular component, $F_{\perp} = F \sin heta$, acts perpendicular to the ground, tending to lift the wagon slightly.
Because the wagon undergoes zero vertical displacement, the perpendicular component does zero mechanical work. Only the parallel component accomplishes work:
W = (F \cos\theta) \times s = F s \cos\theta
Here, $ heta$ represents the angle between the force vector and the displacement vector.
The Three Regimes: Positive, Negative, and Zero Work
Depending on the angle $ heta$ between the force and displacement vectors, mechanical work falls into three distinct physical categories:
+---------------+---------------------+--------------------+-------------------+
| Work Regime | Angle Range | Mathematical Value | Physical Effect |
+---------------+---------------------+--------------------+-------------------+
| Positive Work | 0 <= theta < 90 deg | cos(theta) > 0 | Transfers energy |
| | (Force aids motion) | W > 0 | INTO the object |
| Zero Work | theta = 90 deg | cos(90) = 0 | No mechanical |
| | (Force perp to disp)| W = 0 | energy transfer |
| Negative Work | 90 < theta <= 180 | cos(theta) < 0 | Removes energy |
| | (Force opposes disp)| W < 0 | FROM the object |
+---------------+---------------------+--------------------+-------------------+
The Porter with Luggage: Zero Work Paradox
Normal Upward Lift Force (F)
^
|
+----+----+
| Luggage |
+----+----+
|
=================+================= <-- Porter's Head
|
+-----------------> Horizontal Walk (s)
Angle theta = 90 degrees ---> cos(90 deg) = 0
Work Done by Porter on Luggage = F * s * cos(90) = 0 Joules!
Positive Work ($W > 0$)
When the force has a component in the same direction as the displacement ($0^\circ \le heta < 90^\circ$), work is positive. A horse pulling a plow, an engine accelerating a train, or gravity accelerating a dropped apple performs positive work, adding mechanical energy to the moving body.
Negative Work ($W < 0$)
When the force opposes the direction of displacement ($90^\circ < heta \le 180^\circ$), work is negative. The premier example is kinetic friction: as a box slides across a floor to the right, friction acts to the left ($ heta = 180^\circ$, $\cos 180^\circ = -1$), doing negative work: $W_{ ext{fric}} = -f_k s$. Negative work removes mechanical energy from the object, dissipating it into thermal energy. When you catch a falling ball, your hands exert an upward force while the ball moves downward, performing negative work to absorb the ball's kinetic energy.
Zero Work ($W = 0$)
Work is identically zero under three distinct physical conditions:
1. When displacement is zero ($s = 0$), such as pushing a stationary boulder.
2. When the net force is zero ($F = 0$), such as an asteroid drifting at constant velocity through interstellar space.
3. When the force acts strictly perpendicular to the displacement ($ heta = 90^\circ$). A porter carrying a suitcase on their head walks horizontally forward: the porter exerts an upward vertical lifting force against gravity, but the displacement is horizontal ($ heta = 90^\circ$). Because $\cos 90^\circ = 0$, the vertical force does zero mechanical work on the suitcase. Similarly, in uniform circular motion, centripetal force acts radially inward perpendicular to the tangential displacement, performing zero work and keeping kinetic energy constant.
Work Done by a Variable Force
In real-world mechanics, forces are rarely constant. When stretching a mechanical spring, the resisting force increases linearly with displacement according to Hooke's Law ($F = kx$). When an applied force varies with position, work cannot be calculated by simple multiplication. Instead, work is determined graphically by computing the area beneath the force-displacement curve:
W = \text{Area beneath the } F\text{-}s \text{ curve}
Force (F)
^
F | /|
| / |
| / | Area of Triangle = (1/2) * Base * Height
| / | = (1/2) * x * (kx)
| / | = (1/2) * k * x^2 (Work Done on Spring)
| / |
0 +-------+------+---------------------------> Displacement (x)
0 x
For a linear spring where $F = kx$, the area under the force-displacement line forms a right triangle with base $x$ and height $F = kx$. The work done in stretching the spring from equilibrium to extension $x$ is:
W = \frac{1}{2} \times \text{base} \times \text{height} = \frac{1}{2} x (kx) = \frac{1}{2} k x^2
Chapter Exercises
Exercise 1: A warehouse worker pulls a heavy storage crate across a smooth concrete floor through a distance of 15.0 meters. The worker applies a steady tension force of 120 newtons through a rope inclined at an angle of 30.0 degrees above the horizontal floor. The opposing kinetic frictional resistance between the crate and the floor is a constant 45.0 newtons. Calculate the work done on the crate by the worker's pulling force, calculate the work done by the frictional force, and determine the net work accomplished on the crate.
Exercise 2: A satellite of mass 1200 kilograms orbits Earth in a circular orbit of radius 7000 kilometers at a constant orbital speed of 7.5 kilometers per second. Calculate the work done on the satellite by Earth's gravitational attraction during one complete orbital revolution. Explain your reasoning using vector angles.
Exercise 3: A gardener pushes a lawnmower with a force of 80 newtons directed downward along the handle at an angle of 45 degrees to the horizontal lawn. If the lawnmower is pushed forward through a distance of 25 meters, calculate the work done by the gardener. If the gardener lowers the handle angle to 20 degrees while applying the same 80-newton force over 25 meters, determine whether the work done increases or decreases, and calculate the new value.
Solutions and Explanations
Solution 1: Identify the components of the worker's pulling force: the force is $F = 120 ext{ N}$, displacement is $s = 15.0 ext{ m}$, and angle is $ heta = 30.0^\circ$. The work done by the pulling force is $W_{ ext{worker}} = F s \cos heta = 120 imes 15.0 imes \cos(30.0^\circ) = 1800 imes 0.8660 \approx 1558.8$ joules. For kinetic friction, the force acts directly opposite to the forward displacement, so $ heta = 180^\circ$. The work done by friction is $W_{ ext{friction}} = f_k s \cos(180^\circ) = 45.0 imes 15.0 imes (-1) = -675.0$ joules. The net work done on the crate is the algebraic sum of both work values: $W_{ ext{net}} = W_{ ext{worker}} + W_{ ext{friction}} = 1558.8 + (-675.0) = 883.8$ joules. This positive net work accelerates the crate, increasing its kinetic energy.
Solution 2: In a circular orbit, Earth's gravitational force acts radially inward toward Earth's center, functioning as the centripetal force. The instantaneous displacement of the satellite at every point along its trajectory is directed along the tangent to the circle. Because the radius and tangent of a circle are strictly perpendicular, the angle between the gravitational force vector and the displacement vector is identically ninety degrees ($ heta = 90^\circ$) at every single point along the path. Since $\cos 90^\circ = 0$, the instantaneous work done by gravity is zero ($W = F s \cos 90^\circ = 0$). Over one complete revolution, the gravitational force does exactly 0 joules of work on the satellite. Consequently, the satellite's speed and orbital kinetic energy remain perfectly constant.
Solution 3: In the first case, the force is $F = 80 ext{ N}$, displacement $s = 25 ext{ m}$, and angle $ heta = 45^\circ$. The work done is $W_1 = F s \cos(45^\circ) = 80 imes 25 imes 0.7071 \approx 1414.2$ joules. In the second case, the angle is reduced to $ heta = 20^\circ$. Because the cosine function increases as the angle decreases toward zero ($\cos 20^\circ \approx 0.9397 > \cos 45^\circ \approx 0.7071$), a larger proportion of the applied force aligns horizontally with the direction of displacement, meaning the work done increases. The new work done is $W_2 = 80 imes 25 imes \cos(20^\circ) = 2000 imes 0.9397 \approx 1879.4$ joules. Lowering the handle angle increases the work accomplished on the mower by approximately 465.2 joules for the exact same distance and total applied muscular effort.