Work Done by Variable Forces

1. Introduction

In physics, work is one of the most important quantities that connects force with energy. When a constant force acts on an object and displaces it, calculating the work done is simple: W=F⋅d⋅cos⁡θW = F \cdot d \cdot \cos \thetaW=F⋅d⋅cosθ

But in real life, forces are rarely constant. Think of:

Stretching a spring (force increases with extension).
Gravity changing with altitude.
Friction that depends on velocity.
Rockets expelling gases.

In all these cases, the force varies with position (or sometimes time). That’s where the idea of work done by variable forces comes into play.

This post will guide you step by step through:

Concept of variable forces
Why constant force formulas fail here
The role of calculus in calculating work
Force–displacement graphs
Work done by springs (Hooke’s law)
Gravitational force at varying distances
Solved examples + practice questions
Real-life applications

By the end, you’ll have a crystal-clear understanding of how to handle variable forces in physics.

2. Recap: Work Done by Constant Force

When a constant force FFF acts on an object, displacing it by ddd at an angle θ\thetaθ: W=Fdcos⁡θW = F d \cos \thetaW=Fdcosθ

If force and displacement are in the same direction: W=FdW = FdW=Fd.
If perpendicular: W=0W = 0W=0.
If opposite: W=−FdW = -FdW=−Fd.

This works well for uniform cases, like pushing a box across a smooth floor.

But what if the force changes with position? The simple formula breaks down.

3. Work Done by Variable Force – The Core Idea

If force varies with displacement, we can’t just multiply “force × distance.”

👉 Instead, we divide the displacement into infinitesimally small parts, where the force is approximately constant, calculate small work dWdWdW, and then add them all using integration.

Formula

W=∫x1x2F(x) dxW = \int_{x_1}^{x_2} F(x) \, dxW=∫x1x2F(x)dx

Where:

F(x)F(x)F(x) = force as a function of displacement
x1,x2x_1, x_2x1,x2 = initial and final positions

This is the general formula for work done by a variable force in one dimension.

4. Force–Displacement Graph Method

Another way to visualize this:

Work done = area under the force–displacement graph.

Example:

Constant force → horizontal line → area = rectangle.
Variable force → curve → area under curve.

👉 This graphical approach is very useful in mechanics.

5. Types of Variable Forces

Variable forces can depend on:

Displacement (x): e.g., spring force F=−kxF = -kxF=−kx.
Velocity (v): e.g., air drag F=−kvF = -kvF=−kv.
Time (t): e.g., rocket thrust decreasing with time.
Position in space (r): e.g., gravitational force F=GMmr2F = \frac{GMm}{r^2}F=r2GMm.

We’ll focus mainly on force vs displacement, since that’s the standard in work–energy discussions.

6. Work Done by a Linearly Varying Force

Suppose force increases linearly with displacement: F(x)=axF(x) = axF(x)=ax

Work done: W=∫0dax dx=ad22W = \int_{0}^{d} ax \, dx = a \frac{d^2}{2}W=∫0daxdx=a2d2

👉 Area under a straight-line graph = triangle.

7. Work Done by a Spring (Hooke’s Law)

Hooke’s Law

For a spring: F(x)=−kxF(x) = -kxF(x)=−kx

Where:

kkk = spring constant
xxx = extension or compression

Work Done

W=∫0xF(x) dx=∫0x(−kx)dxW = \int_{0}^{x} F(x) \, dx = \int_{0}^{x} (-kx) dxW=∫0xF(x)dx=∫0x(−kx)dx W=−12kx2W = -\frac{1}{2} k x^2W=−21kx2

👉 Negative sign indicates work done by the spring.
👉 Work done on the spring = +12kx2+\frac{1}{2}kx^2+21kx2.

This is exactly the elastic potential energy stored in a spring.

8. Work Done by Gravitational Force (Variable)

Close to Earth’s surface, gravity is constant (F=mgF = mgF=mg).
But at large distances: F(r)=GMmr2F(r) = \frac{GMm}{r^2}F(r)=r2GMm

Work done moving from r1r_1r1 to r2r_2r2: W=∫r1r2GMmr2drW = \int_{r_1}^{r_2} \frac{GMm}{r^2} drW=∫r1r2r2GMmdr W=GMm(1r1−1r2)W = GMm \left(\frac{1}{r_1} – \frac{1}{r_2}\right)W=GMm(r11−r21)

👉 This leads to the concept of gravitational potential energy: U(r)=−GMmrU(r) = -\frac{GMm}{r}U(r)=−rGMm

9. Work Done by Non-Linear Forces

Sometimes force varies as a power of displacement:

F(x)=ax2⇒W=a3(x3)F(x) = ax^2 \Rightarrow W = \frac{a}{3}(x^3)F(x)=ax2⇒W=3a(x3)
F(x)=a/x2⇒W=−axF(x) = a/x^2 \Rightarrow W = -\frac{a}{x}F(x)=a/x2⇒W=−xa

Each case requires integration based on the function of force.

10. Work–Energy Theorem with Variable Forces

Even with variable forces, the work-energy theorem still holds: Wnet=ΔKW_{net} = \Delta KWnet=ΔK

That means the total work done (calculated using integration) equals the change in kinetic energy of the body.

11. Real-Life Applications

Stretching Springs – storing energy in trampolines, bows, car suspensions.
Planetary Motion – work by varying gravity on satellites.
Roller Coasters – variable forces as the slope changes.
Bungee Jumping – elastic potential energy in cords.
Atomic Physics – electrostatic force varies with distance.

12. Solved Examples

Example 1: Linearly Increasing Force

A force increases with displacement: F=2xF = 2xF=2x (N), where xxx is in meters. Find work done from x=0x = 0x=0 to x=4 mx = 4 \, mx=4m. W=∫042x dx=[x2]04=16 JW = \int_{0}^{4} 2x \, dx = [x^2]_0^4 = 16 \, JW=∫042xdx=[x2]04=16J

Example 2: Spring Work

A spring constant k=200 N/mk = 200 \, N/mk=200N/m is stretched by 0.1 m. Find work done. W=12kx2=12(200)(0.1)2=1 JW = \frac{1}{2} k x^2 = \frac{1}{2}(200)(0.1)^2 = 1 \, JW=21kx2=21(200)(0.1)2=1J

Example 3: Gravitational Force

Find work done to move a satellite of mass 1000 kg1000 \, kg1000kg from Earth’s surface (R=6.4×106mR = 6.4 \times 10^6 mR=6.4×106m) to double that distance. Take GM=6.67×10−11×6×1024=3.986×1014GM = 6.67 \times 10^{-11} \times 6 \times 10^{24} = 3.986 \times 10^{14}GM=6.67×10−11×6×1024=3.986×1014. W=GMm(1R−12R)W = GMm \left(\frac{1}{R} – \frac{1}{2R}\right)W=GMm(R1−2R1) =GMm2R= \frac{GMm}{2R}=2RGMm =3.986×1014×10002×6.4×106= \frac{3.986 \times 10^{14} \times 1000}{2 \times 6.4 \times 10^6}=2×6.4×1063.986×1014×1000 =3.1×1010 J= 3.1 \times 10^{10} \, J=3.1×1010J

13. Common Misconceptions

“Work requires force only.” → No, displacement must also occur.
“Variable forces can be handled with average values.” → Not always, only integration gives exact result.
“Springs store work directly.” → They store energy as potential, which equals work done on them.

14. Practice Problems

A force F=5xF = 5xF=5x acts on a body. Find work from x=0x = 0x=0 to x=3 mx = 3 \, mx=3m.
Work done in stretching a spring of constant 100 N/m by 20 cm.
A force F=10/xF = 10/xF=10/x acts from x=1x = 1x=1 to x=4x = 4x=4. Find work.
Calculate work done by gravity in moving a 5 kg mass from Earth’s surface to 3R.
Explain why area under F-x graph represents work.

15. Conclusion

The concept of work done by variable forces bridges simple mechanics and advanced physics. Unlike constant forces, variable forces require calculus: W=∫F(x)dxW = \int F(x) dxW=∫F(x)dx

Key takeaways:

Work is area under force-displacement graph.
Springs and gravity are classic examples.
Even with variable forces, the work-energy theorem holds true.
These principles explain satellites, roller coasters, rockets, and even atomic physics.

So next time you stretch a spring or watch a rocket launch, remember – it’s not magic, it’s the mathematics of variable forces at work!