- #1
physicN00Bz
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Homework Statement
A block is being launch of a ramp, the length of the ramp is 5m(hypotenuse). The ramp also has a kinetic coefficient of .2 and the block has an initial velocity of 12m/s.
Homework Equations
-Wx-uFn = ma
Velocity equations for finding distance off the ramp
The Attempt at a Solution
ok to find the velocity at the end of the ramp.
-mgsin35-umgcos(35) = ma
15.6 - 1.6 = -7.2 s
V^2 + Vi^2 + 2at(change in x)
V= Square root(12^2 + 2(-7.2)(5))
V at the end of the ramp is 8.47 ms (this was confirm right by the answer sheets)
finding the velocity in all direction
Vy = 4.86 ms
Vx = 6.94 ms
V = 8.47 ms
now for the distance, I am going to find the time it takes for Vy to reach zero
Vyf = Vyi +at
0 = 4.8 + (-9.8)t
t=.49
time it takes for it to fall down
Height of the ramp 2.87
Height of the pojectable when Vy = 0 2.384
total height 5.25
0= 5.25 + 1/2(-9.8)t
T= 1.07s
Total time = 1.57s
x=0+6.94(1.57) = 10.9m
but the actual answer is 9.78
why am i off?