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.
![P=\frac{W}{t}](https://tex.z-dn.net/?f=P%3D%5Cfrac%7BW%7D%7Bt%7D)
The work is the force the is applicated in a distance so
W=F*d
replacing:
and
determinate distance in time is velocity so
a).
![P=F*v](https://tex.z-dn.net/?f=P%3DF%2Av)
![F=m*a\\F=m*g*sin(33.5)](https://tex.z-dn.net/?f=F%3Dm%2Aa%5C%5CF%3Dm%2Ag%2Asin%2833.5%29)
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
![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}}](https://tex.z-dn.net/?f=vi%3D0%5Cfrac%7Bm%7D%7Bs%7D%20%5C%5Cvf%3D2.15%20%5Cfrac%7Bm%7D%7Bs%7D%5C%5Cvf%3Dvi%2Ba%2At%5C%5Cvf-vi%3Da%2At%5C%5C%20a%3D%5Cfrac%7Bvf-vi%7D%7Bt%7D%3D%20a%3D%5Cfrac%7B2.15%5Cfrac%7Bm%7D%7Bs%7D-0%5Cfrac%7Bm%7D%7Bs%7D%7D%7B13s%7D%5C%5Ca%3D0.1653%20%5Cfrac%7Bm%7D%7Bs%5E%7B2%7D%7D)
The maximum force is when the car is accelerating so
![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](https://tex.z-dn.net/?f=Ft%3DFa%2BFg%5C%5CFt%3Dm%2Aa%2Bm%2Ag%2Asin%2833.5%29%5C%5CFt%3D950kg%2A0.1653%5Cfrac%7Bm%7D%7Bs%5E%7B2%7D%7D%2B950%2A9.8%5Cfrac%7Bm%7D%7Bs%5E%7B2%7D%7D%2Asin%2833.5%29%5C%5CFt%3D5295.565%20N)
so the maximum force is the maximum force by the maximum speed
![Pmax=Ft*v\\Pmax=5295.565N*2.15\frac{m}{s}\\Pmax=11385.46\\Pmax=11.3854kW](https://tex.z-dn.net/?f=Pmax%3DFt%2Av%5C%5CPmax%3D5295.565N%2A2.15%5Cfrac%7Bm%7D%7Bs%7D%5C%5CPmax%3D11385.46%5C%5CPmax%3D11.3854kW)
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