A 1380 kg car starts from rest at the top of a 28 m long hill inclined at 11.0 degrees. Ignoring friction, how fast is it going when it reaches the bottom of the hill?
How much Mechanical Energy does it start with?
(f/i=final/initial)
1/2mvf^2+mghf=1/2mv^2i+mghi
+37091 F=ma
the force DOWN the plane (parallel with surface) = 1380 (9.81) sin 11 = 2583.13 N
F=ma
2583.13 N = 1380 kg (a)
a=1.872 m/s^2
x = xo +vot + 1/2 a t^2 And realize xo = 0 vo = 0
x= 28m = 1/2 (1.872) t^2
t = 5.47 sec
vf= at = 1.872 (5.47) = 10.24 m/s
Hope I didn't make a mistake!
mgh = potential energy = 1380 (9.81)28 sin11= 72327.75J
As a check (and a simpler method) the POTENTIAL energy is converted to KINETIC energy at the bottom of the incline
Kinetic energy = 1/2mv^2 Potential energy = kinetic energy
72327.75 = 1/2 (1380) v^2 v= 10.24 m/s Which agrees with what we found (the long way) above !
