Question

Determine the power required for a 1150-kg car to climb a 100-m-long uphill road with a slope of 30° (from horizontal) in 12 s (a) at a constant velocity, (b) from rest to a final velocity of 30 m/s, and (c) from 35 m/s to a final velocity of 5 m/s. Disregard friction, air drag, and rolling resistance.

Answer 1

A) 14487 W

B) 38445 W

C) -23813 W

Step-by-step explanation:

A) Constant velocity (a=0)

Force balance

`F-P=m*a`

`F-m*g*cos(30)=0`

`F=1150kg*9.8m/s^2*cos(30)=1738.4 N`

Energy:
`E=F*d=1738.4 N*100m=173840 J`

Power:
`Power=E/t=173840 J/12 s=14487 W`

B) Aceleration:
`a=((30-0)m/s)/(12s)=2.5m/s^2`

Force balance

`F-P=m*a`

`F-m*g*cos(30)=1150kg*2.5m/s^2`

`F=2875N + 1150kg*9.8m/s^2*cos(30)=4613.4 N`

Energy:
`E=F*d=4613.4 N*100m=461340 J`

Power:
`Power=E/t=461340 J/12 s=38445 W`

C) Aceleration:
`a=((5-35)m/s)/(12s)=-2.5m/s^2`

Force balance

`F-P=m*a`

`F-m*g*cos(30)=1150kg*-2.5m/s^2`

`F=-2875N + 1150kg*9.8m/s^2*cos(30)=-1136.6 N`

Energy:
`E=F*d=-1136.6 N*100m=-113660 J`

Power:
`Power=E/t=113660 J/12 s=-23813 W`

Answer 2

(a) 46.95 kW

(b) 90.1 kW

(c) 104.46 kW

Step-by-step explanation:

L = 100 m , θ = 0 degree, m = 1150 kg, t = 12 s

(a) When it is moving with constant velocity

Energy, E = m x g x h = m x g x L x Sinθ = 1150 x 9.8 x 100 x Sin 30 =

= 563500 J

Power, P = Energy / Time = 563500 / 12 = 46958.3 watt = 46.95 kW

(b) From rest to final velocity of 30 m/s

E = m x g x L Sinθ + 1/2 mv^2

E = 1150 x 9.8 x 100 x Sin 30 + 0.5 x 1150 x 30 x 30 = 1081000 J

P = E / T = 1081000 / 12 = 90083.33 W = 90.1 kW

(c) from 35 m/s to 5 m/s

E = m x g x L Sinθ + 1/2 mv^2 - 1/2 mu^2

E = 1150 x 9.8 x 100 x Sin 30 + 0.5 x 1150 x 35 x 35 - 0.5 x 1150 x 5 x 5

E = 563500 + 704375 - 14375 = 1253500 J

P = E / T = 1253500 / 12 = 104458.33 W = 104.46 kW