Question

Please find the power (in kW) needed in accelerating a 1000 kg car from 0 to 100 km/hour in 10 seconds on a 5% gradient up-hill road.

Answer 1

P= 45.384 kW

Step-by-step explanation:

given data:

m = 1000 kg

u = 0

v = 100 km/hr = 250/9 m/s

t = 10 sec

g =9.81 m/s2

5% gradient

from figure we can have

`tan\theta = (0.05x)/(x)`

`\theta =  tan^(-1)0.05`

`\theta = 2.86` from equation of motion we have

v = u + at

`(250)/(9) = 0 + 9*10`

`a = (25)/(9) m/s^2`

distance covered in 10 sec

from equation of motion

`s = ut + (1)/(2)at^2`

`s = 0*10+ (1)/(2)*(25)/(9)*10^2`

s = 138.8 m

from newton's 2nd law of motion along inclined position

`F -mgsin\theta  = ma`

solving for f

`f = mgsin\theta +ma`

`F = 1000*9.81*SIN2.8624 +1000*(25)/(9)`

F = 3267.67 N

work done is given as W

`W = F* s`

and power
`P = (W)/(t)`

`P = (F*s)/(t)`

`P = (3267.67*(1250)/(9))/(10)`

P = 45384.30 W

P= 45.384 kW