IP Combined Science Notes (Lower Sec, Year 1-2): 08) Work, Energy, Power & Pressure

Energy never disappears — it changes form. Track these transformations with equations, careful unit usage, and thought-out assumptions.

Learning targets

• Calculate work done, kinetic energy, and gravitational potential energy in SI units.
• Relate power to rate of energy conversion and discuss efficiency improvements.
• Apply pressure formulas in solids (force over area) and fluids (depth dependence).
• Interpret hydraulic system diagrams and solve transmission-of-force problems.

1. Work & energy

Assume \( \theta = 0^\circ \) for force parallel to displacement unless stated otherwise.

Worked example — Roller coaster drop

A \( \pu{450 kg} \) car descends \( \pu{25 m} \).

1. Loss in \( E_p \): \( 450 \times 9.81 \times 25 = 1.10 \times 10^{5} , \pu{J} \).
2. Neglecting losses, this becomes kinetic energy. Solve for \( v \):

\[ \tfrac{1}{2} \times 450 \times v^2 = 1.10 \times 10^{5} \Rightarrow v = 22.2 , \pu{m.s-1}. \]

1. Discuss real-world losses (air resistance, friction) reducing final speed.

2. Power & efficiency

\[ P = \frac{W}{t} = Fv. \]

Efficiency:

\[ \eta = \frac{\text{Useful output energy}}{\text{Input energy}} \times 100%. \]

Appliance example

An electric kettle rated \( \pu{2.4 kW} \) heats \( \pu{1.2 kg} \) of water from \( \pu{22 ^\circ C} \) to \( \pu{100 ^\circ C} \) in \( \pu{4.0 min} \).

1. Energy absorbed by water:

\[ Q = mc\Delta T = 1.2 \times 4.18 \times (100 - 22) = 392 , \pu{kJ}. \]

1. Electrical energy input:

\[ E = P t = 2.4 \times 10^{3} \times (4.0 \times 60) = 576 , \pu{kJ}. \]

1. Efficiency \( = \frac{392}{576} \times 100 = 68.1% \).

3. Pressure in solids

\[ P = \frac{F}{A}. \]

Example — Snowshoes

A person of mass \( \pu{68 kg} \) stands on snowshoes with total area \( \pu{0.32 m2} \).

\[ F = mg = 68 \times 9.81 = 667 , \pu{N}. \]

\[ P = \frac{667}{0.32} = 2.08 \times 10^{3} , \pu{Pa}. \]

Compare with regular shoes area \( \pu{0.08 m2} \) to highlight fourfold increase in pressure.

4. Pressure in fluids

Hydrostatic pressure at depth \( h \):

\[ P = \rho g h. \]

Example: Water density \( \pu{1000 kg.m-3} \), depth \( \pu{1.5 m} \):

\[ P = 1000 \times 9.81 \times 1.5 = 1.47 \times 10^{4} , \pu{Pa}. \]

Hydraulic lift

If master piston area \( \pu{50 cm2} \) and slave piston area \( \pu{500 cm2} \):

• Apply \( \pu{120 N} \) on master piston.
• Pressure transmitted equally: \( P = \frac{120}{50 \times 10^{-4}} = 2.40 \times 10^{4} , \pu{Pa} \).
• Output force: \( F = P \times A = 2.40 \times 10^{4} \times 500 \times 10^{-4} = 1200 , \pu{N} \).

Mechanical advantage of 10 achieved assuming negligible losses.

Try it yourself

1. A \( \pu{65 kg} \) hiker climbs \( \pu{420 m} \) in \( \pu{35 min} \). Determine change in gravitational potential energy and average useful power output.
2. Two people push a crate with force \( \pu{210 N} \) over \( \pu{12 m} \), with \( \pu{5%} \) of work lost to heat. Calculate useful work delivered to the crate.
3. A hydraulic press has a large piston diameter twice that of the small piston. If \( \pu{80 N} \) is applied on the small piston, find the output force (ignore losses) and describe two practical limitations of this model.

Next, study waves at https://eclatinstitute.sg/blog/ip-combined-sciences-lower-sec-notes/IP-Combined-Science-Lower-Sec-09-Waves-Light-and-Sound.