Kinetic energy and the work-energy theorem
Energy of motion
A fast cricket ball can knock the stumps flying. A moving body can do work on whatever it hits. We say it has energy.
The energy a body has because it is moving is called kinetic energy. We write it as K.
For a body of mass m moving with speed v, K equals one half m v squared.
A 160 gram cricket ball at 30 metres per second: half of 0.16 times 900. That is 72 joules.
Watch two balls side by side. One has twice the mass: double the kinetic energy. The other has twice the speed: four times the kinetic energy.
Speed counts double because it is squared. That is why a car at 80 km/h is far more dangerous in a crash than one at 40.
Kinetic energy is a scalar. It is never negative, and its unit is the joule, just like work.
The work-energy theorem
Here is a block of mass m on a smooth floor, moving with speed u. A steady force F pushes it through a distance d.
From kinematics we already know: v squared minus u squared equals 2 a d.
Multiply both sides by m over 2. On the left: half m v squared minus half m u squared. That is the change in kinetic energy.
On the right: m a times d. But m a is the net force F, so this is F times d, the work done.
So the change in kinetic energy equals the work done by the net force. This is the work-energy theorem.
Positive work speeds a body up. Negative work, like braking, slows it down. Zero work leaves its speed unchanged.
The theorem works even when the force changes from point to point. To see why, we need a graph.
Work done by a variable force
A constant force on a force–position graph is a flat line. The work F times d is the area of the rectangle underneath.
Now let the force change as the body moves. The graph becomes a curve.
Chop the motion into thin strips of width Δx. Across one strip the force hardly changes, so the work is F times Δx, a thin rectangle.
Add up all the strips. As the strips get thinner, the jagged steps fit the curve exactly.
So the work done by a variable force is the area under the force–position graph. In calculus, it is the integral of F dx.
Where the curve dips below the axis, the force points backwards, and that area counts as negative work.
Add the areas, and you get the change in kinetic energy, just as before.
Try it: work changes kinetic energy
- The floor is smooth, so the only horizontal force is F cos θ.
- Press Release and compare “Work by F” with “Change in KE”. They always match.
- Set θ above 90° to make the force slow the block down.
Quick check
A 5 kg block slides along a smooth floor at 4 m/s. A force acting opposite to its motion brings it to rest in 5 m. How much work does this force do?
- +40 J
- −40 J
- −20 J
- −8 J
Answer
B. By the work-energy theorem, W = Kf − Ki = 0 − ½ × 5 × 4² = −40 J. The work is negative because the force points against the motion. (The force itself is 40 J ÷ 5 m = 8 N.)
Area under the graph
A force along the x-axis varies as F = 5x newtons (x in metres). How much work does it do as the body moves from x = 0 to x = 2 m?
- 20 J
- 5 J
- 10 J
- 2.5 J
Answer
C. The F–x graph is a straight line from 0 N at x = 0 to 10 N at x = 2 m. The area under it is a triangle: ½ × 2 m × 10 N = 10 J. Multiplying 10 N by 2 m (20 J) would wrongly assume the force was 10 N the whole way.
Summary
- Kinetic energy:
K = ½mv² - Work-energy theorem:
K_f − K_i = W_net - Work by a variable force:
W = ∫ F dx (area under the F–x graph)
- Kinetic energy depends on v²: double the speed and K becomes four times bigger.
- The change in kinetic energy equals the work done by the net force.
- Positive work speeds a body up, negative work slows it down.
- For a changing force, the work is the area under the force–position graph.