Students will learn how to calculate power and efficiency. Students will also learn the true meaning of both.

### Key Equations

\begin{align*}P = \frac{W} {\Delta t}\end{align*} ; Power is equal to the energy released per second and has units of Watts ( 1 Watt = J/s).

\begin{align*}\text{Eff} = \frac{P_{out}}{P_{in}}\end{align*} ; The efficiency equals the output power divided by the input power

#### Example 1

**Question**: A pile driver lifts a 500 kg mass a vertical distance of 20 m in 1.1 sec. It uses 225 kW of supplied power to do this.

a) How much work was done by the pile driver?

b) How much power was used in actually lifting the mass?

c) What is the efficiency of the machine? (This is the ratio of power used to power supplied.)

d) The mass is dropped on a pile and falls 20 m. If it loses 40,000 J on the way down to the ground due to air resistance, what is its speed when it hits the pile?

**Answer**:

a) We will use the equation for work (which gives us the amount of energy transferred) and plug in the known values to get the amount of work done by the pile driver. \begin{align*} W=Fd=mgd=500\mathrm{kg}\times 9.8\mathrm{m/s^2}\times 20\mathrm{m}=9.8\times 10^4\mathrm{J} \end{align*}

b) We will use the power equation and plug in the known values and then convert to kW at the end. \begin{align*} P=\frac{W}{\Delta t}=\frac{9.8\times 10^4\mathrm{J}}{1.1\mathrm{s}}=89000\mathrm{W} \times \frac{1 \mathrm{kW}}{1000\mathrm{W}}=89\mathrm{kW} \end{align*}

c) Efficiency is defined as the Power out divided by the Power in. Thus, this is simply a division problem. \begin{align*} Eff = \frac{\mathrm{power~used}}{\mathrm{power~supplied}}=\frac{89\mathrm{kW}}{225\mathrm{kW}}=.40 \end{align*}

d) We have already solved for the amount of energy the mass has after the pile driver performs work on it (it has \begin{align*}9.8\times 10^4\mathrm{J}\end{align*}). If on the way down it loses \begin{align*}40000\mathrm{J}\end{align*} due to air resistance, then it effectively has \begin{align*} 98000\mathrm{J}-40000\mathrm{J}=58000\mathrm{J} \end{align*} of energy. So we will set the kinetic energy equation equal to the total energy and solve for v. This will give us the velocity of the mass when it hits the ground because right before the mass hits the ground, all of the potential energy will have been converted into kinetic energy. \begin{align*} 58000\mathrm{J}=\frac{1}{2}mv^2 \Rightarrow v=\sqrt{\frac{58000\mathrm{J}\times 2}{m}}=\sqrt{\frac{58000\mathrm{J}\times 2}{500\mathrm{kg}}}=15.2\mathrm{m/s} \end{align*}

### Watch this Explanation

### Simulation

### Explore More

- Before a run, you eat an apple with 1,000,000 Joules of binding energy.
- 550,000 Joules of binding energy are wasted during digestion. How much remains?
- Some 95% of the remaining energy is used for the basic processes in your body (which is why you can warm a bed at night!). How much is available for running?
- Let’s say that, when you run, you lose 25% of your energy overcoming friction and air resistance. How much is available for conversion to kinetic energy?
- Let’s say your mass is 75 kg. What could be your top speed under these idealized circumstances?
- But only 10% of the available energy goes to KE, another 50% goes into heat exhaust from your body. Now you come upon a hill if the remaining energy is converted to gravitational potential energy. How high do you climb before running out of energy completely?

- A pile driver’s motor expends 310,000 Joules of energy to lift a 5400 kg mass. The motor is rated at an efficiency of 0.13. How high is the mass lifted?
- 15. A 1500 kg car starts at rest and speeds up to 3.0 m/s with a constant acceleration.
- What is the car’s gain in kinetic energy?
- What power is exerted by the engine?
- We define efficiency as the ratio of output energy (in this case kinetic energy) to input energy. If this car’s efficiency is 0.30, how much input energy was provided by the gasoline?
- If 0.00015 gallons were used up in the process, what is the energy content of the gasoline in Joules per gallon?
- Compare that energy to the food energy in a gallon of Coke, if a 12-oz can contains 150 Calories (food calories) and one gallon is 128 ounces.

#### Answers to Selected Problems

- a. 450,000 J b. 22,500 J c. 5,625 J d. 21.2 m/s e. 9.18 m
- 0.76 m
- a. 6750 J b. 5.6 kW c. 22.5 kJ d. 150 MJ/gallon of gas e. 6.7 MJ/gallon Coke