Answer:
a). P=11.04kW
b). Pmax=11.38 kW
c). Wt=6423.166kJ
Explanation:
The power of the motor when the speed is constant is the work in a determinate time.
[tex]P=\frac{W}{t}[/tex]
The work is the force the is applicated in a distance so
W=F*d
replacing:
[tex]P=F*\frac{d}{t}[/tex] and [tex]\frac{d}{t}[/tex] determinate distance in time is velocity so
a).
[tex]P=F*v[/tex]
[tex]F=m*a\\F=m*g*sin(33.5)[/tex]
[tex]P=950kg*9.8\frac{m}{s^{2}}*sin(33.5)*2.15\frac{m}{s}\\P=11047.846 W\\P=11.0478 kW[/tex]
b).
The maximum power must the motor provide, is the maximum force with the maximum speed of the motor in this case
The first step is find the acceleration so
[tex]vi=0\frac{m}{s} \\vf=2.15 \frac{m}{s}\\vf=vi+a*t\\vf-vi=a*t\\ a=\frac{vf-vi}{t}= a=\frac{2.15\frac{m}{s}-0\frac{m}{s}}{13s}\\a=0.1653 \frac{m}{s^{2}}[/tex]
The maximum force is when the car is accelerating so
[tex]Ft=Fa+Fg\\Ft=m*a+m*g*sin(33.5)\\Ft=950kg*0.1653\frac{m}{s^{2}}+950*9.8\frac{m}{s^{2}}*sin(33.5)\\Ft=5295.565 N[/tex]
so the maximum force is the maximum force by the maximum speed
[tex]Pmax=Ft*v\\Pmax=5295.565N*2.15\frac{m}{s}\\Pmax=11385.46\\Pmax=11.3854kW[/tex]
c).
The total energy transfer without any friction is the weight move in the high axis y in this case, so is easy to know that distance
W=m*g*h
h=Length*sin(33.5)
W=m*g*Length*sin(33.5)
W=950 kg*9.8* 1250m*sin(33.5)
W=6423166.667 kJ
W=6423.166 kJ
