Work, Energy and Power -_-"

[anthropos]

Yo, i am a 15-year-old student. One day I saw this in my assignment.

 

A car of mass 1500 kg accelerates from rest at 5.00 m/s2 for 5.00 s on a straight road. Find teh minimum power needed for the engine to achieve this task.

 

The answer they gave was

 

Power = work done/time

= 1/2 mv2 / t

=1/2 m(u+at)2 / t

=1/2(1500)(0+5x5)^2 / 5

=93750

=93.8W

 

But my answer was

 

Since m=1500 kg

u=0

a=5.00 m/s2

t=5.00s

F=ma=1500x 5.00 = 7500N

v=u +at =5.00 x 5 =25m/s

P=Fv ==> 7500x25 = 187500 = 187.5 kW

 

Not only the answer is wrong, 187500 is twice as 93750. What happened? Please enlighten me! Thanks.

[hybrid04]

Well im not quite sure where you went with this P=Fv it may be correct i havent dont one of these problems in a while, but i think that might be the source. W=Fd; You can figure out d based on acceleration and time and F from F=ma. So you have F and d. Power by definition is Work per unit time so you divide W by the given amount of time for P.

 

The way they did it was pretty similar they instead of using d just found final velocity and calculated the kinetic energy then divided it by time for power.

 

It seems to me like you have the general idea just mixed up some definition of terms somewhere. The fact that it is twice that might have to do with the 1/2mv^2 or just pure coincidence. Id just relook at how you defined things and try to understand my methodology and you should be good.

[swansont]

But my answer was

 

Since m=1500 kg

u=0

a=5.00 m/s2

t=5.00s

F=ma=1500x 5.00 = 7500N

v=u +at =5.00 x 5 =25m/s

P=Fv ==> 7500x25 = 187500 = 187.5 kW

 

Not only the answer is wrong' date=' 187500 is twice as 93750. What happened? Please enlighten me! Thanks. [/quote']

 

Was the car going at 25 m/s the whole time? Your answer gives how much power was being generated at t= 5s, but at t=0, the power generated was zero.

 

If you think about how the final speed relates to the average speed I think you'll see why you're off by a factor of two.