Power (Grade 12 NSC Matric Physical Sciences): Revision Notes

Worked Example 1: Moving a box

Question: Calculate the power required for a force of 10 N applied to move a 10 kg box at a speed of 1 m/s over a frictionless surface.

Solution:

• Given: $F = 10 \text{ N}$, $v = 1 \text{ m/s}$
• Required: Power

Using $P = F \times v$: $P = (10 \text{ N}) \times (1 \text{ m/s}) = 10 \text{ W}$

Answer: 10 W of power are required to move the box at constant speed.

Worked Example 2: Forklift lifting a crate

Question: A forklift lifts a crate of mass 100 kg to a height of 8 m in 4 s, then holds it in place for 20 s. Calculate the power during each phase.

Solution:

Phase 1 - Lifting the crate:

• Mass = 100 kg, height = 8 m, time = 4 s
• Force needed = weight = $mg = (100 \text{ kg})(9.8 \text{ m/s}^2) = 980 \text{ N}$
• Displacement = 8 m
• Velocity = $\frac{8 \text{ m}}{4 \text{ s}} = 2 \text{ m/s}$

$P = F \times v = (980 \text{ N}) \times (2 \text{ m/s}) = 1960 \text{ W}$

Phase 2 - Holding the crate: When holding the crate in place, there is no displacement, so no work is done. $P = 0 \text{ W}$

Answer: 1960 W to lift the crate, 0 W to hold it in place.

Worked Example 3: Climbing stairs

Question: What is the power output for a 60.0 kg woman who runs up a 3.00 m high flight of stairs in 3.50 s, starting from rest but having a final speed of 2.00 m/s?

Solution: The total work done includes both gravitational potential energy and kinetic energy:

$W = \frac{1}{2}mv^2 + mgh$ $W = \frac{1}{2}(60.0)(2.00)^2 + (60.0)(9.8)(3.00)$ $W = 120 + 1764 = 1884 \text{ J}$

$P = \frac{W}{t} = \frac{1884 \text{ J}}{3.50 \text{ s}} = 538 \text{ W}$

Answer: The woman generates 538 W of power.

Worked Example 4: Pumping water

Question: What is the power required to pump water from a borehole with depth h = 15.0 m at a rate of 20.0 l/s?

Solution:

• Volume per second = 20.0 l/s = 0.020 m³/s
• Mass of water per second = 20.0 kg/s
• Height = 15.0 m
• Velocity at surface = 15.0 m/s

The work includes both potential and kinetic energy: $W = \frac{1}{2}mv^2 + mgh$ $W = \frac{1}{2}(20.0)(15.0)^2 + (20.0)(9.8)(15.0)$ $W = 2250 + 2940 = 5190 \text{ J}$

$P = \frac{W}{t} = \frac{5190 \text{ J}}{1 \text{ s}} = 5.19 \times 10^3 \text{ W}$

Answer: The minimum power required is $5.19 \times 10^3$ W.