# How much work can a 22 kW car engine do in 60 seconds if it is: a. 100% efficient b. 50%...

## Question:

How much work can a 22 kW car engine do in 60 seconds if it is:

a. 100% efficient

b. 50% efficient

c. 15% efficient

## Work, Energy, and Efficiency:

Mechanical power is the rate of doing work. The product of power and time gives the work done. The efficiency (usually denoted by {eq}\mu {/eq} ) of doing work (given per unit or as a percentage) is the rate of conversion of the input energy Pi into useful work Po.

{eq}\begin{align*} \mu &= \frac{P_o}{P_i} \end{align*} {/eq}

Given the power P = 22 kW , the time t = 60 s and the efficiency {eq}\mu {/eq}, the work done is given by the equation {eq}\text{Work Done} = \mu Pt {/eq}.

#### Part a.

When {eq}\mu = 100\ \% {/eq},

{eq}\begin{align*} \text{Work Done} &= \mu Pt\\ &= 100\ \% \ \cdot(22\ \text{kW})(60\ \text{s})\\ &\color{blue}{= \boxed{1.32\ \times 10^6\ \text{J}}} \end{align*} {/eq}

#### Part b.

When {eq}\mu = 50\ \% {/eq},

{eq}\begin{align*} \text{Work Done} &= \mu Pt\\ &= 100\ \% \ \cdot(22\ \text{kW})(60\ \text{s})\\ &\color{blue}{= \boxed{6.6\ \times 10^5\ \text{J}}} \end{align*} {/eq}

#### Part c.

When {eq}\mu = 15\ \% {/eq},

{eq}\begin{align*} \text{Work Done} &= \mu Pt\\ &= 15\ \% \ \cdot(22\ \text{kW})(60\ \text{s})\\ &\color{blue}{= \boxed{1.98\ \times 10^5\ \text{J}}} \end{align*} {/eq} 