7.4 Power (Page 2/8)

University physics volume 1 Page 2 / 8

Check Your Understanding Estimate the power expended by a weightlifter raising a 150-kg barbell 2 m in 3 s.

980 W

The power involved in moving a body can also be expressed in terms of the forces acting on it. If a force $\vec{F}$ acts on a body that is displaced $d \vec{r}$ in a time dt , the power expended by the force is

P = \frac{d W}{d t} = \frac{\vec{F} \cdot d \vec{r}}{d t} = \vec{F} \cdot (\frac{d \vec{r}}{d t}) = \vec{F} \cdot \vec{v},

where $\vec{v}$ is the velocity of the body. The fact that the limits implied by the derivatives exist, for the motion of a real body, justifies the rearrangement of the infinitesimals.

Automotive power driving uphill

How much power must an automobile engine expend to move a 1200-kg car up a 15% grade at 90 km/h ( [link] )? Assume that 25% of this power is dissipated overcoming air resistance and friction.

An automobile is shown moving up along a 15 percent grade at a speed of v = 90 kilometers per hour. The car has mass m = 1200 kilograms. — We want to calculate the power needed to move a car up a hill at constant speed.

Strategy

At constant velocity, there is no change in kinetic energy, so the net work done to move the car is zero. Therefore the power supplied by the engine to move the car equals the power expended against gravity and air resistance. By assumption, 75% of the power is supplied against gravity, which equals

m \vec{g} \cdot \vec{v} = m g v sin θ,

where

θ

is the angle of the incline. A 15% grade means

tan θ = 0.15 .

This reasoning allows us to solve for the power required.

Solution

Carrying out the suggested steps, we find

0.75 P = m g v sin ({tan}^{−1} 0.15),

P = \frac{(1200 \times 9.8 N) (90 m / 3.6 s) sin (8.53 °)}{0.75} = 58 kW,

or about 78 hp. (You should supply the steps used to convert units.)

Significance

This is a reasonable amount of power for the engine of a small to mid-size car to supply

(1 hp = 0.746 kW).

Note that this is only the power expended to move the car. Much of the engine’s power goes elsewhere, for example, into waste heat. That’s why cars need radiators. Any remaining power could be used for acceleration, or to operate the car’s accessories.

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Summary

Power is the rate of doing work; that is, the derivative of work with respect to time.
Alternatively, the work done, during a time interval, is the integral of the power supplied over the time interval.
The power delivered by a force, acting on a moving particle, is the dot product of the force and the particle’s velocity.

Key equations

Work done by a force over an infinitesimal displacement	$d W = \vec{F} \cdot d \vec{r} = \| \vec{F} \| \| d \vec{r} \| cos θ$
Work done by a force acting along a path from A to B	$W_{A B} = \int_{path A B} \vec{F} \cdot d \vec{r}$
Work done by a constant force of kinetic friction	$W_{fr} = − f_{k} \| l_{A B} \|$
Work done going from A to B by Earth’s gravity, near its surface	$W_{grav, A B} = − m g (y_{B} - y_{A})$
Work done going from A to B by one-dimensional spring force	$W_{spring, A B} = − (\frac{1}{2} k) (x_{B}^{2} - x_{A}^{2})$
Kinetic energy of a non-relativistic particle	$K = \frac{1}{2} m v^{2} = \frac{p^{2}}{2 m}$
Work-energy theorem	$W_{net} = K_{B} - K_{A}$
Power as rate of doing work	$P = \frac{d W}{d t}$
Power as the dot product of force and velocity	$P = \vec{F} \cdot \vec{v}$

Conceptual questions

Most electrical appliances are rated in watts. Does this rating depend on how long the appliance is on? (When off, it is a zero-watt device.) Explain in terms of the definition of power.

Appliances are rated in terms of the energy consumed in a relatively small time interval. It does not matter how long the appliance is on, only the rate of change of energy per unit time.

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Source: OpenStax, University physics volume 1. OpenStax CNX. Sep 19, 2016 Download for free at http://cnx.org/content/col12031/1.5

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