Answer
a) $P=47.01kW$
b) $P=90.14kW$
c) $P=-10.49kW$
Work Step by Step
$h=100m*sin(30^{\circ})=50m$
The work of climb is:
$W_{p}=mg\Delta z=1150kg*9.81\frac{m}{s^2}*50m=564.08kJ$
And the power of climb is:
$P_{p}=\frac{W_{p}}{t}=\frac{564.08kJ}{12s}=47.01kW$
a) In this case, since velocity is constant the power needed is only the one to climb the uphill road:
$P_{t,a}=P_{p}=47.01kW$
b) In this case, the power needed is the sum of the one to climb the uphill road and the one to increase the velocity:
$W_{k,b}=\frac{1}{2}m\Delta v^2=\frac{1}{2}*1150kg*((30\frac{m}{s})^2-(0\frac{m}{s})^2)=517.5kJ$
$P_{k,b}=\frac{W_{k,b}}{t}=\frac{517.5kJ}{12s}=43.13kW$
Then the total power needed is:
$P_{t,b}=P_{k,b}+P_{p}=43.13kW+47.01kW=90.14kW$
c) In this case, the power needed is the sum of the one to climb the uphill road and the one to reduce the velocity:
$W_{k,c}=\frac{1}{2}m\Delta v^2=\frac{1}{2}*1150kg*((5\frac{m}{s})^2-(35\frac{m}{s})^2)=-690kJ$
$P_{k,c}=\frac{W_{k,c}}{t}=\frac{-690kJ}{12s}=-57.5kW$
Then the total power needed is:
$P_{t,c}=P_{k,c}+P_{p}=-57.5kW+47.01kW=-10.49kW$
