solve problems using P = W / t

Energy Conservation

In a closed system the total energy remains constant. Energy can be transformed from one form to another (e.g. kinetic ↔ potential, chemical ↔ thermal) but the sum of all forms does not change.

Key Definitions

• Work (W): The transfer of energy when a force moves an object through a distance. $W = F \times s \times \cos\theta$ (Joules).
• Power (P): The rate at which work is done or energy is transferred. $P = \dfrac{W}{t}$ (Watts).
• Energy (E): The capacity to do work. Common forms include kinetic energy $E_k = \frac12 mv^2$, gravitational potential energy $E_p = mgh$, and elastic potential energy $E_e = \frac12 kx^2$.

Units

Quantity	Symbol	SI Unit	Unit Symbol
Work / Energy	$W$, $E$	Joule	J
Power	$P$	Watt	W
Force	$F$	Newton	N
Distance	$s$	Metre	m
Time	$t$	Second	s

Using $P = \dfrac{W}{t}$ to Solve Problems

When a problem gives any two of the three quantities (power, work/energy, time), the third can be found directly from the definition of power.

Step‑by‑step approach

1. Identify what is asked: power, work, or time.
2. Write down the known quantities with correct units.
3. Convert all quantities to SI units before substituting.
4. Use $P = \dfrac{W}{t}$ and rearrange as needed:
   • To find work: $W = P \times t$
   • To find time: $t = \dfrac{W}{P}$
   • To find power: $P = \dfrac{W}{t}$
5. Check that the result is consistent with the principle of energy conservation (e.g., total energy before = total energy after).

Example Problem 1

Question: A 1500 W electric heater raises the temperature of 2.0 kg of water from 20 °C to 80 °C. How long does it take? (Specific heat capacity of water $c = 4180\ \text{J kg}^{-1}\text{K}^{-1}$.)

Solution:

1. Calculate the energy required (work done on the water): $$W = mc\Delta T = (2.0\ \text{kg})(4180\ \text{J kg}^{-1}\text{K}^{-1})(80-20\ \text{K}) = 5.02\times10^{5}\ \text{J}$$
2. Rearrange $P = \dfrac{W}{t}$ to find time: $$t = \dfrac{W}{P} = \dfrac{5.02\times10^{5}\ \text{J}}{1500\ \text{W}} = 334\ \text{s}$$
3. Convert to minutes if desired: $334\ \text{s} \approx 5.6\ \text{min}$.

Example Problem 2

Question: A cyclist climbs a 300 m hill at a constant speed of 3 m s⁻¹. The combined mass of cyclist and bike is 80 kg. Assuming no losses, what is the average power output of the cyclist?

Solution:

1. Work done against gravity (increase in gravitational potential energy): $$W = mgh = (80\ \text{kg})(9.81\ \text{m s}^{-2})(300\ \text{m}) = 2.36\times10^{5}\ \text{J}$$
2. Time taken to climb: $$t = \dfrac{\text{distance}}{\text{speed}} = \dfrac{300\ \text{m}}{3\ \text{m s}^{-1}} = 100\ \text{s}$$
3. Power: $$P = \dfrac{W}{t} = \dfrac{2.36\times10^{5}\ \text{J}}{100\ \text{s}} = 2360\ \text{W}$$

Common Pitfalls

• Forgetting to convert minutes or hours to seconds before using $P = W/t$.
• Mixing up energy forms: work done by non‑conservative forces (e.g., friction) must be included in the energy balance.
• Assuming 100 % efficiency; in real systems, account for losses by introducing an efficiency factor $\eta$: $P_{\text{useful}} = \eta P_{\text{input}}$.

Linking Power to Energy Conservation

Energy conservation states that the total energy change $\Delta E_{\text{total}}$ of a system is zero when no energy crosses the system boundary. Power provides a convenient way to track how quickly that energy change occurs:

$$\sum P_{\text{in}} - \sum P_{\text{out}} = \dfrac{dE_{\text{total}}}{dt} = 0 \quad (\text{steady state})$$

Thus, in steady‑state problems the sum of all power inputs equals the sum of all power outputs.