6.11 Power

WHAT IS POWER?

Power is defined as the time rate at which work is done or energy is transferred. The average power of a force is defined as the ratio of the work, \(W\), to the total time \(t\) taken

\(P_{a v}=\frac{W}{t}\)
The instantaneous power is defined as the limiting value of the average power as the time interval approaches zero,
\(P=\frac{\mathrm{d} W}{\mathrm{~d} t} \dots  (6.5)\)
The work \(d W\) done by a force \(F\) for a displacement \(d \vec{r}\) is \(\mathrm{d} W=\vec{F}\). \(d \vec{r}\). The instantaneous power can also be expressed as
\(
\begin{aligned}
P &=\vec{F} \cdot \frac{\mathrm{d} \vec{r}}{\mathrm{d} t} \\
&=\vec{F} \cdot \vec{v}
\end{aligned}
\)
where \(\vec{v}\) is the instantaneous velocity when the force is \(\vec{F}\).

Power, like work and energy, is a scalar quantity. Its dimensions are [ML \(\left.{ }^{2} \mathrm{~T}^{-3}\right]\). In the SI, its unit is called a watt (W). The watt is \(1 \mathrm{~J} \mathrm{~s}^{-1}\). The unit of power is named after James Watt, one of the innovators of the steam engine in the eighteenth century. There is another unit of power, namely the horsepower (hp)
\(
1 \mathrm{hp}=746 \mathrm{~W}
\)

Example 6.16:

An elevator can carry a maximum load of \(1800 \mathrm{~kg}\) (elevator + passengers) is moving up with a constant speed of \(2 \mathrm{~m} \mathrm{~s}^{-1}\). The frictional force opposing the motion is \(4000 \mathrm{~N}\). Determine the minimum power delivered by the motor to the elevator in watts as well as in horsepower.

Solution: The downward force on the elevator is
\(F=m g+F_{f}=(1800 \times 10)+4000=22000 \mathrm{~N}\)
The motor must supply enough power to balance this force. Hence,
\(P=\vec{F} . \vec{v}=22000 \times 2=44000 \mathrm{~W}=59 \mathrm{hp}\)

Example 6.17:

Consider a drop of mass \(1.00 \mathrm{~g}\) falling from a height \(1.00 \mathrm{~km}\). It hits the ground with a speed of \(50.0 \mathrm{~m} \mathrm{~s}^{-1}\). (a) What is the work done by the gravitational force? What is the work done by the unknown resistive force?

Solution: The change in kinetic energy of the drop is
\(
\begin{array}{l}
\Delta K=\frac{1}{2} m v^{2}-0 \\
=\frac{1}{2} \times 10^{-3} \times 50 \times 50 \\
=1.25 \mathrm{~J}
\end{array}
\)
where we have assumed that the drop is initially at rest.

Assuming that \(g\) is a constant with a value \(10 \mathrm{~m} / \mathrm{s}^{2}\), the work done by the gravitational force is,
\(
\begin{aligned}
W_{g} &=m g h \\
&=10^{-3} \times 10 \times 10^{3} \\
&=10.0 \mathrm{~J}
\end{aligned}
\)
(b) From the work-energy theorem
\(
\Delta K=W_{g}+W_{r}
\)
where \(W_{r}\) is the work done by the resistive force on the raindrop. Thus
\(
\begin{aligned}
W_{r} &=\Delta K-W_{g} \\
&=1.25-10 \\
&=-8.75 \mathrm{~J}
\end{aligned}
\)
is negative.