Interactive physics simulator
Work: W = Fs
Explore the fundamental definition of work. Decompose constant force paths, analyze work against gravity, and integrate the area under Force-Displacement graphs.
Work Equation Lab (W = Fs)
Trace the mechanical work done by forces and study the area under the Force-Displacement curve in real time.
Live Telemetry
- Applied Force (F)
- 50.0 N
- Displacement (s)
- 8.0 m
- Work Formula
- W = F · s
- Work Done (W)
- 400.0 J
Understanding Mechanical Work
In physics, work measures the energy transferred by a force acting on an object as it moves. Work is done only when there is a displacement along the line of action of the force. If the force acts parallel to the motion, the work done is simply the force magnitude multiplied by the distance traveled: W = F · s.
Key Principles
For work to be done, two conditions must be satisfied:
- Force Application: A net external force must act on the body. Pushing against a stationary wall does zero work since the wall does not move.
- Displacement: The body must move a distance s (d). If there is no displacement (s = 0), the work done is exactly zero.
- Directional Alignment: Only the force parallel to motion does work. Perpendicular forces do zero work.
Constant vs Variable Force
Calculating work depends on whether the force remains constant or changes over the distance:
- Constant Force (W = Fs): Force magnitude is constant. On an F-s graph, this is a rectangle of height F and width s.
- Variable Force (W = Area): The force changes (e.g. stretching a spring requires more force as it stretches, F=kx). The work done equals the triangular area under the curve: W = (1)/(2) k x2.
Solved Examples
A constant horizontal pulling force of F = 45 N is applied to drag a wooden crate across a friction-free warehouse floor. If the crate is moved through a straight-line displacement of s = 14 meters in the exact direction of the force, calculate the work done on the crate.
- Identify the given variables: Force magnitude F = 45 N, displacement s = 14 m.
- Identify the angle between force and displacement: Since the force is applied in the exact direction of motion, the angle θ = 0°, so cos(0°) = 1.
- Recall the simplified formula for work: W = F · s.
- Substitute the values into the equation: W = 45 N · 14 m.
- Calculate the result: W = 630 Joules.
- The pulling force performs 630 J of work, transferring 630 J of kinetic energy to the crate.
Answer: Work Done W = 630 J
A construction hoist lifts a steel beam of mass m = 25 kg vertically upward to a height of h = 8 meters. Calculate the work done by the hoist against gravity. (Use gravity acceleration g = 9.8 m/s²).
- Identify the given values: Mass m = 25 kg, vertical height/displacement h = 8 m, g = 9.8 m/s².
- Calculate the upward pulling force required to lift the beam at a constant speed: F = Fg = m · g = 25 · 9.8 = 245 N.
- Recall the vertical work formula (work done against gravity): W = F · h = m · g · h.
- Substitute values: W = 245 N · 8 m = 1960 Joules.
- The hoist performs 1,960 J of work to raise the beam, increasing its gravitational potential energy by 1.96 kJ.
Answer: Work Done W = 1960 J (1.96 kJ)
An elastic metal spring with a spring constant of k = 150 N/m is stretched from its relaxed position by a distance of x = 0.4 meters. Calculate the work done to stretch the spring. Explain how this is represented on a force-displacement graph.
- Identify the variables: Spring constant k = 150 N/m, stretch distance x = 0.4 m.
- Recall Hooke's Law for variable force: F = k · x. The force is not constant; it increases linearly from 0 N to Fmax = 150 · 0.4 = 60 N.
- The work done is equal to the area of the triangle under the Force-Displacement graph: W = 1/2 · Fmax · x = 1/2 · k · x².
- Substitute values into the spring work equation: W = 1/2 · 150 · (0.4)² = 75 · 0.16 = 12 Joules.
- The work done to stretch the spring is 12 J, which is stored as elastic potential energy.
Answer: Work Done W = 12 J
Common Mistakes
- Assuming effort equals work. Holding a heavy weight still takes muscular effort, but does zero physical work because displacement is zero.
- Forgetting units conversion. Ensure mass is in kg (then multiply by 9.8 to find force in Newtons) and displacement is in meters.
- Ignoring variable force profiles. Using W = Fmax · s for a spring instead of integrating the area under the triangular Hooke's Law slope.
Work-Energy Theorem
The net work done on a particle equals the change in its kinetic energy. Doing positive work increases its velocity, while negative work extracts energy, slowing it down.
Practice Questions
1. A student pushes a heavy brick wall with a force of F = 150 N for 2 minutes, but the wall does not move. How much work is performed on the wall?
Since the wall does not move, its displacement is exactly s = 0. According to the formula W = F · s, since s = 0, the work done on the wall is W = 150 N · 0 m = 0 J. Although the student exerts effort and burns physiological energy, no mechanical work is done on the wall in the physical sense.
2. If you double the applied force on an object but move it half the original distance, how does the work done change?
Work is proportional to both force and displacement (W = F · s). If you replace F with 2F, and s with 0.5s, the new work done is W' = (2F) · (0.5s) = F · s. Therefore, the work done remains exactly the same as the original value.
3. What does the area under a Force vs Displacement graph represent in physics?
The area under a Force vs Displacement (F-s) graph represents the total mechanical work done by the force. For a constant force, the area is a rectangle (height F, width s). For a variable force like a spring, the area is a triangle or a complex shape, representing the integral of force over the distance.
4. Why is the work done by gravity on a satellite in a perfect circular orbit equal to zero?
In a perfect circular orbit, the gravitational force (which acts radially inward toward the planet) is always perpendicular (at an angle of 90°) to the satellite's instantaneous displacement vector (which points tangentially along the orbit). Since cos(90°) = 0, gravity does zero work on the satellite, and its speed remains constant.
FAQ
Frequently Asked Questions
What is the definition of work in physics?
Work is the transfer of energy that occurs when a force acts on an object, causing it to move a distance in the direction of the force.
What is the formula for work done?
The basic formula is W = F · s, where W is work in Joules, F is the magnitude of force parallel to motion in Newtons, and s is the displacement in meters.
What is the SI unit of work?
The SI unit of work is the Joule (J), which equals one Newton-meter (1 N·m).
Can work be negative?
Yes. Work is negative when the applied force opposes the direction of displacement (e.g. friction forces acting to slow down a sliding box).
What is the difference between positive, negative, and zero work?
Positive work adds energy to an object (force assists motion). Negative work extracts energy (force opposes motion). Zero work does not transfer energy (no displacement, or force is perpendicular to motion).
How do you calculate work done against gravity?
Work done against gravity is calculated as W = m · g · h, where m is mass (kg), g is gravity (9.8 m/s²), and h is the vertical height lifted (meters).
What is a Joule equivalent to in base units?
In SI base units, one Joule (J) is equivalent to 1 kg·m²/s².
Why is no work done when you carry a heavy suitcase horizontally at a constant speed?
The upward lifting force you exert to support the suitcase is vertical, while the displacement is horizontal. Because the force and displacement vectors are perpendicular (90°), the carrying force does zero work on the suitcase.
How do you find work done from a Force-Displacement graph?
The total work done is equal to the geometric area under the force line on a Force vs Displacement graph.
What is Hooke's Law and how does it relate to work?
Hooke's Law states that spring force is proportional to stretch (F = kx). Since the force varies linearly, the work done to stretch a spring is the area of a triangle: W = 1/2 · k · x².
What is the Work-Energy Theorem?
The Work-Energy Theorem states that the net work done on an object is equal to the change in its kinetic energy (W_net = ΔKE).
Does holding a heavy barbell stationary above your head do work?
No. Because the barbell remains stationary, its displacement is zero (s = 0), meaning the physical work done on the barbell is exactly zero.