Work, Energy and PowerNotions of Work and Kinetic Energy: The Work-Energy Theorem
Kabir has built a go-kart for the school fair, and Sina has bagged the driver's seat. Kabir can push hard for a short distance or gently for a long one. Which push sends Sina off faster? In this chapter blue marks the force that does the work and the kinetic energy it creates, green marks displacements and velocities, purple marks the result (the work, the total energy), and amber marks friction and the heat it makes.
Choose a force and a distance, then push. The cart's mass is 2 kg, and the track is smooth.
Every push drew a rectangle on the graph, force times distance. The blue area always matched the cart's kinetic energy at the end, ½mv², and two very different pushes with the same area gave exactly the same speed. That product of force and distance is the work done on the cart.
Where ½mv² Comes From
For motion in a straight line with constant acceleration a, Chapter 2 gave
where u and v are the initial and final speeds and s is the distance covered. Multiply both sides by m/2 and use Newton's second law, F = ma:
In three dimensions, with an acceleration vector a and a displacement vector d, Eq. (5.2) becomes , and multiplying by m/2 again,
The left side is the change in "half the mass times the speed squared", which we call the kinetic energy K. The right side is the displacement times the part of the force along it, which we call the work W. So
Work is always done by a particular force, on a particular body, over a particular displacement.
Kabir pushes the 2 kg go-kart from rest with 9 N over 2 m. The work done is
The work-energy theorem: the change in the kinetic energy of a particle equals the work done on it by the net force.
½ × 2 kg × v² = 18 J gives v² = 18 m²/s², so v = 4.24 m/s. We have proved the theorem only for a constant force; Section 5.6 extends it to forces that change from place to place.
A Raindrop and the Air
Example 5.2 A falling raindrop is pulled down by gravity and held back by air resistance, a force that grows with the drop's speed but whose exact form we do not know. A drop of mass 1.50 g falls from a cloud 800 m up and reaches the ground at 40 m/s. Take g = 10 m/s². (a) How much work does gravity do on the drop? (b) How much work does the air resistance do?
Solution: The drop starts from rest, so its change in kinetic energy is
=
(a) Gravity is a constant force mg acting along the 800 m fall:
=
(b) By the work-energy theorem, , so
=
Gravity put in the purple 12 J, the air took back the amber 10.8 J, and the drop kept only the blue 1.2 J as kinetic energy. The work done by the air is negative, and we found it without knowing the law of air resistance at all: the theorem only needs the speeds at the start and the end.