CALL THE SLOPE ANGLE theta.
We can get the slope angle from the slope S (given as value 1/5) using arctangent.
theta = arctan(S)
We will desire to access all trigonometry of theta...so that is why we desire to express as an angle instead of a slope.
Forces perpendicular to the road (define up as +y):
Component of gravity: -m*g*cos(theta)
Normal force: N
These forces add up to zero, because the road is assumed straight, and thus no acceleration normal to the road.
Hence:
N = m*g*cos(theta)
Forces parallel to the road (define forward as +x):
Component of weight: -m*g*sin(theta)
Drag: -D
Traction: +F
Newton's second law for these forces:
F - m*g*sin(theta) - D = m*a
Solve for F:
F = m*(a + g*sin(theta)) + D
NOW...we substitute previous expressions for a and theta:
a = (vf^2 - vi^2)/(2*d)
theta = arctan(S)
Thus:
F = m*((vf^2 - vi^2)/(2*d) + g*sin(arctan(S))) + D
We can simplify the trig(arctrig) combination as:
sin(arctan(S)) = S/sqrt(1 + S^2)
Thus:
F = m*((vf^2 - vi^2)/(2*d) + g*S/sqrt(1 + S^2) ) + D
Since we also desire work done, we want to evaluate:
W = integral (F dot dx)
OR...since force is constant (special case):
W = F*d
Thus:
W = d*(m*((vf^2 - vi^2)/(2*d) + g*S/sqrt(1 + S^2) ) + D)
To get POWER, we also need the TIME INTERVAL.
t = 2*d*(vf - vi)/(vf^2 - vi^2)
By definition of AVERAGE power:
P_avg = W/t
Thus:
P_avg = d*(m*((vf^2 - vi^2)/(2*d) + g*S/sqrt(1 + S^2) ) + D)/(2*d*(vf - vi)/(vf^2 - vi^2))
Simplify:
P_avg = (m*((vf^2 - vi^2)/(2*d) + g*S/sqrt(1 + S^2) ) + D)/(2*(vf - vi)/(vf^2 - vi^2))
P_avg = (m*((vf^2 - vi^2)/(2*d) + g*S/sqrt(1 + S^2) ) + D)*(vf^2 - vi^2)/(2*(vf - vi))
Summary:
A: F = m*((vf^2 - vi^2)/(2*d) + g*S/sqrt(1 + S^2) ) + D
B: W = d*(m*((vf^2 - vi^2)/(2*d) + g*S/sqrt(1 + S^2) ) + D)
C: P_avg = (m*((vf^2 - vi^2)/(2*d) + g*S/sqrt(1 + S^2) ) + D)*(vf^2 - vi^2)/(2*(vf - vi))
DATA:
m = 1350 kg
g = 9.8 Newtons/kilogram; PLEASE...remember where this came from. It is not an arbitrary number...it has physical meaning as Earth's gravitational field.
d = 100 m
S = 0.2
vi = 2.5 m/s
vf = 12 m/s
D = 750 Newtons
Results:
A: F = 4275 Newtons
B: W = 42750 Joules
C: P_avg = 26074 Watts
D: you didn't ask for it, but it is important when sizing engines. Maximum power: P_max = F*v_max ... thus P_max = F*vf
P_max = vf*(m*((vf^2 - vi^2)/(2*d) + g*S/sqrt(1 + S^2) ) + D)
P_max = 51294 Watts