Answer:
47 KW
90.1 KW
-10.5 KW
Explanation:
Knowns and requirements
m=1150 kg—> mass of the car
L=100 m—> road length
It's required to determine the power needed to climb the uphill road (30° from horizontal ) in 12 second
a) at constant speed
b) from rest to a final velocity of 30 m/s
c) from 35 m/s to a final velocity of 5 m/s
a) The power required to climb the uphill road can simply expressed as the rate of change of the kinetic energy and potential energy
P=m*g*Δh/Δt+0.5m(v_2^2-v_1^2)/Δt (1)
where Δh is the vertical rise of the road (Δh = L sin(30))
Since the car moving at constant speed, there will be no change in kinetic energy and Eq.(1) reduced to
P=m*g*Δh/Δt
=1150*9.8*100*sin(30)/12s
=47 KW
b) Bulging our values (v_1 = 0.0) and (v_2 = 30 m/s) into Eq.(1), The required power will be
P=m*g*Δh/Δt+0.5m(v_2^2-v_1^2)/Δt
=90.1 KW
c) Bulging our values (v_1= 35) and (v_2 = 5 mis) into Eq.(1), The required power will be
P=m*g*Δh/Δt+0.5m(v_2^2-v_1^2)/Δt
=-10.5 KW
Note : The negative sign means that the car is losing energy. In other words the driver must use the brake to reduce the velocity from 35 mis to 5 mis during climbing the uphill road.
