8.2. Power http://www.ck12.orgSolution:The force exerted must be equal to the weight of the man:mg= ( 70 .0 kg)( 9 .80 m/s^2 ) =686 NW=F d= (686 N)( 4 .5 m) =3090 N m=3090 J
P=Wt =^30904. 0 sJ=770 J/s=770 W
P=770 W= 1 .03 hp
SinceP=Wt andW=F d, we can use these formulas to derive a formula relating power to the speed of the object
that is produced by the power.
P=Wt =F dt =Fdt=F v
The velocity in this formula is the average speed of the object during the time interval.
Example Problem:Calculate the power required of a 1400 kg car if the car climbs a 10° hill at a steady 80. km/h.
Solution:First convert 80. km/h to m/s: 22.2 m/s.
In 1.00 s, the car would travel 22.2 m on the road surface but the distance traveled upward would be (22.2 m)(sin 10°) =
(22.2 m)(0.174) = 3.86 m. The force in the direction of the upward motion is the weight of the car: (1400 kg)(9.80 m/s^2 )
= 13720 N.W=F d= (13720 N)( 3 .86 m) = 53 ,000 J
Since this work was done in 1.00 second, the power would be 53,000 W.
This problem can be solved a different way; by calculating the upward component of the velocity of the car. The
process would be similar, and start with finding the vertical component of the velocity vector: (22.2 m/s)(sin 10°) =
(22.2 m/s)(0.174) = 3.86 m/s. Again, calculate the weight of the car: (1400 kg)(9.80 m/s^2 ) = 13720 N. Finally, we
could use the formula relating power to average speed to calculate power.
P=F v= (13720 N)( 3 .86 m/s) = 53 ,000 WSummary- Power is defined as the rate at which work is done or the rate at which energy is transformed.
- Power=WorkTime
- Power=Force×velocity
PracticeQuestions
Use the video below to answer the following questions about work and power.
https://www.youtube.com/watch?v=u6y2RPQw7E0MEDIA
Click image to the left for use the URL below.
URL: http://www.ck12.org/flx/render/embeddedobject/78703- What is the difference between positive and negative work?
- What are the standard units for power?
- What is horsepower?