The energy way is the easy way but.
Normal force = m g cos 14
force down ramp = m g sin 14 + mu m g cos 14 = m g (mu cos 14+ sin 14)
= m g ( 1.12 )
F = m a
a = F/m = g (1.12) = 10.9 m/s^2 down ramp
say t is time to stop
v = Vi - 10.9 t
0 = Vi - 10.9 t
so
t = Vi/10.9
d = Vi t - (10.9/2) t^2
85 = Vi^2/10.9 - (10.9/2)(Vi/10.9)^2
85 = (Vi^2/10.9 )( 1 - 1/2)
170 = Vi^2/10.9
Vi = 43 m/s
A car with engine trouble slides 85 meters up along a 14 degree ramp before coming to a stop. If the coefficient of friction between the tires and the road is 0.9, what was the velocity of the car at the bottom of the ramp?
I'm not looking for a solution which includes Potential Energy. We haven't learned that in class yet, is there a another way to solve this?
1 answer
3 answers