Work and Energy Notes

Calculate work, kinetic energy, gravitational potential energy, and power for simple cases.
Apply conservation of energy and identify energy transformations.
Describe the difference between physiological and physical work and recognize the limits of human power and energy outputs.

The work done by a constant force F is given by the equation: W = Fd\cos\theta, where d is the distance moved and \,\theta is the angle between F and d.
This is essentially the distance multiplied by the component of the force in the direction of displacement.
If \,\theta = 90^\circ, then W = 0.
Example: Pushing a trolley with F = 50 N at 30^\circ to the horizontal, over a distance of d = 4 m.
- Work done on the trolley: W = (50 \, N)(4 \, m)\cos(30^\circ) = 173 \, J

Holding a heavy object may cause muscle ache, which feels like hard work.
However, no physical work is done because the displacement is zero.
Muscles under tension require energy to continually re-trigger muscle fiber contractions; this is physiological work.

Efficiency is the ratio of useful output (physical work) to the metabolic energy used.
Example: Stacking supermarket shelves involves physical work, and the metabolic energy used can be measured.
Efficiency is typically between 5-20%, depending on the muscles used.
\text{Efficiency} = \frac{\text{work out}}{\text{energy used}}

Kinetic energy (KE) of a moving object:
- KE = \frac{1}{2}mv^2
Example: A 60 kg person sprinting at 9 m/s.
- KE = \frac{1}{2}(60 \, kg)(9 \, m/s)^2 = 2430 \, J
If the only energy change is KE when a force does work, then the change in KE is equal to the work done on the object.
- \Delta(KE) = W = \text{work done on object}
- This is known as the work-kinetic energy principle.
Example: What average force operating over 20 m causes a runner to gain a KE of 2430 J?
- \Delta(KE) = W = Fd
- F = \frac{\Delta(KE)}{d} = \frac{2430 \, J}{20 \, m} = 121.5 \, N
The energy source is not the ground pushing the runner.

What constant net force F{\text{net}} is required to accelerate a bus from speed v1 to speed v_2 over a distance d?
- \Delta(KE) = \frac{1}{2}mv2^2 - \frac{1}{2}mv1^2
- \Delta(KE) = F{\text{net}}d = \frac{1}{2}mv2^2 - \frac{1}{2}mv1^2 = \frac{1}{2}m(v2^2 - v_1^2)
- F{\text{net}} = \frac{m(v2^2 - v_1^2)}{2d}

In raising a book of mass m through a height h, the work done by the external force F_{\text{ext}} (by hand) is:
- W = F_{\text{ext}}d = mgh
An object is said to have gravitational potential energy due to its height y.
- GPE = mgy
We use “gravitational PE” rather than “work by gravitational force”.
The change in the book’s gravitational PE is mgy2 - mgy1 = mgh

If an object falls freely due to gravity, gravitational potential energy decreases, and kinetic energy increases.
The total energy, (KE + GPE), is constant.
If the roller-coaster is frictionless, then you have the same KE at the end that you started with if you end up at height y

Power is the time rate of transferring energy (e.g., doing work, heating water, metabolizing carbohydrates).
Unit: J \cdot s^{-1} or Watt (W).
Example: A 60 kg sprinter goes from 0 to 10 m/s and acquires 3000 J of kinetic energy in 4.0 s. Find the average power (energy per second being transformed into kinetic energy).
- Power = \frac{\text{energy}}{\text{time}} = \frac{3000 \, J}{4.0 \, s} = 750 \, W
Typical max human output power is around 500 – 900 W in activities such as cycling and climbing stairs (with relatively high efficiency of converting metabolic energy using large muscles) – higher for good athletes.
P = \frac{\Delta}{\Delta t} \text{(energy)} which is the rate of transferring energy

Most biomechanical processes are limited by either:
- Power – ability to deliver a lot of energy in a short time.
- Total energy to be delivered over a relatively long time.
What is the limitation for each of the following?
- Sprinting short distance.
- Cycling uphill as fast as possible.
- Lifting a very heavy barbell.
- Light barbell lifted with 100 repetitions.
- Climbing up a mountain, 2000 m gain in altitude.