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.

Respuesta :

Answer:

(a) 46.95 kW

(b) 90.1 kW

(c) 104.46 kW

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

Answer:

A) 14487 W

B) 38445 W

C) -23813 W

Explanation:

A) Constant velocity (a=0)

Force balance

[tex]F-P=m*a[/tex]

[tex]F-m*g*cos(30)=0[/tex]

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

Energy: [tex]E=F*d=1738.4 N*100m=173840 J[/tex]

Power: [tex]Power=E/t=173840 J/12 s=14487 W[/tex]

B) Aceleration: [tex]a=\frac{(30-0)m/s}{12s}=2.5m/s^2[/tex]

Force balance

[tex]F-P=m*a[/tex]

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

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

Energy: [tex]E=F*d=4613.4 N*100m=461340 J[/tex]

Power: [tex]Power=E/t=461340 J/12 s=38445 W[/tex]

C) Aceleration: [tex]a=\frac{(5-35)m/s}{12s}=-2.5m/s^2[/tex]

Force balance

[tex]F-P=m*a[/tex]

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

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

Energy: [tex]E=F*d=-1136.6 N*100m=-113660 J[/tex]

Power: [tex]Power=E/t=113660 J/12 s=-23813 W[/tex]