Work–Energy Principle

T_{1} + U_{1 \to 2} = T_{2}, T = \frac{1}{2} m v^{2}

The work of all forces acting on a particle between 1 and 2 equals the change in kinetic energy. Conservative forces may be moved into a potential V so that T1 + V1 = T2 + V2 if no non-conservative work occurs.

$T$ (J): Kinetic energy
$U$ (J): Work from 1 to 2

Step-by-step solved example

A 2 kg collar slides from rest down a 4 m frictionless drop. Speed at the bottom?

Work is the area under F–s, equal to the change in kinetic energy.

1. Energy
ΔV = mgh = 2×9.81×4 = 78.5 J becomes T = ½mv².
$v = 2 g h = 78.5 = 8.86 m/s$

Answer: 8.86 m/s

10 practice questions

0/10 correct

1.KE of 4 kg at 5 m/s is

2.Work of a constant 30 N force through 2 m (aligned) is

3.A 1 kg mass dropped 5 m: speed (g = 9.81)

4.Power is

5.Spring work from 0 to x is

6.If friction does −40 J and ΔT = +10 J, other work is

7.Potential of weight relative to a datum is

8.A 800 kg pile hammer drops 2 m onto a pile that settles 40 mm. Average impact force ≈

9.Work of a force perpendicular to displacement is

10.Conservation of mechanical energy requires