Respuesta :
Answer:
a) [tex]F = 882.63\,N[/tex], b) [tex]\dot W= 4413.15\,W[/tex], c) [tex]\eta = 15.216\,\%[/tex].
Explanation:
a) Let assume that car travel on a horizontal surface. The equations of equilibrium of the car are:
[tex]\Sigma F_{x} = F - \mu_{r}\cdot N = 0[/tex]
[tex]\Sigma F_{y} = N - m\cdot g = 0[/tex]
After some algebraic handling, the following expression for the propulsion force is constructed:
[tex]F = \mu_{r}\cdot m \cdot g[/tex]
[tex]F = (0.06)\cdot (1500\,kg)\cdot (9.807\,\frac{m}{s^{2}} )[/tex]
[tex]F = 882.63\,N[/tex]
b) The power require to move the car at a speed of 5 meters per second is:
[tex]\dot W = F\cdot v[/tex]
[tex]\dot W = (882.63\,N)\cdot (5\,\frac{m}{s} )[/tex]
[tex]\dot W= 4413.15\,W[/tex]
c) The efficiency of the car is:
[tex]\eta = \frac{(882.63\,N)\cdot (15\,mi)\cdot (\frac{1609\,m}{1\,mi} )}{(1.4\times 10^{8}\,J)} \times 100\,\%[/tex]
[tex]\eta = 15.216\,\%[/tex]
The answer to the given question would be as follows:
a). Propulsion Force [tex]= 882.63 N[/tex]
b). Power necessary for Propulsion [tex]= 4413.15 W[/tex]
c). Car's efficiency [tex]= 15.216[/tex]%
a). Propulsion Force:
Given,
Weight of car = 1500 Kg
Speed = 5.0 m/s
Coefficient of rolling friction = 0.06
[tex]F =[/tex] μ[tex]_{r}[/tex] × [tex]m[/tex] × [tex]g[/tex]
[tex]F =[/tex] [tex](0.06) (1500) (9.807 m/s^2)[/tex]
∵ [tex]F = 882.63 N[/tex]
b). Power necessary for Propulsion:
Given that,
Propulsion = 5 m/s
[tex]W = F[/tex] × [tex]v[/tex]
[tex]= (882.63 N) (5 m/s)[/tex]
[tex]= 4413.15 W[/tex]
Now,
c). Capacity of the gallon = [tex]1.4[/tex] × [tex]10^8 J[/tex]
1 mile = 1609 m
η [tex]= {(883.63) (15) (1609)}/{(1.4 * 10^8 J)} * 100[/tex]
∵ η [tex]= 15.216[/tex]%
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