Answer:
P= 45.384 kW
Explanation:
given data:
m = 1000 kg
u = 0
v = 100 km/hr = 250/9 m/s
t = 10 sec
g =9.81 m/s2
5% gradient
from figure we can have
[tex]tan\theta = \frac{0.05x}{x}[/tex]
[tex]\theta = tan^{-1}0.05 [/tex]
[tex]\theta = 2.86[/tex] from equation of motion we have
v = u + at
[tex]\frac{250}{9} = 0 + 9*10[/tex]
[tex]a = \frac{25}{9} m/s^2[/tex]
distance covered in 10 sec
from equation of motion
[tex]s = ut + \frac{1}{2}at^2[/tex]
[tex]s = 0*10+ \frac{1}{2}*\frac{25}{9}*10^2[/tex]
s = 138.8 m
from newton's 2nd law of motion along inclined position
[tex]F -mgsin\theta = ma[/tex]
solving for f
[tex]f = mgsin\theta +ma[/tex]
[tex]F = 1000*9.81*SIN2.8624 +1000*\frac{25}{9}[/tex]
F = 3267.67 N
work done is given as W
[tex]W = F* s[/tex]
and power [tex]P = \frac{W}{t}[/tex]
[tex]P = \frac{F*s}{t}[/tex]
[tex]P = \frac{3267.67*\frac{1250}{9}}{10}[/tex]
P = 45384.30 W
P= 45.384 kW
