A 2.0 kg wood block is launched up a wooden ramp that is inclined at a 30∘ angle. The block's initial speed is 10m/s. The coefficient of kinetic friction of wood on wood is μk=.200.
What vertical height does the block reach above its starting point?
The answer is 3.79m, but I didn't arrive at it using my method.
Looking it up on google, I saw people talking about conservation of energy to answer it, but we aren't at that point in this class yet. We're working on friction.
The Attempt at a Solution
m = 2 kg
mu_k = 0.200
v_0 = 10 m/s
theta = 30 degrees
First I found the net x and y forces:
F_net_x = f_k + F_G*sin(30 degrees)
F_net_y = F_N + F_G*cos(30 degrees)
Normal force = F_G*cos(30 degrees)*(-1) = 16.97
f_k = 0.200*16.97 = 3.39
So acceleration for the block is -1.7 (a = F/m, negative because going in the -x direction.)
Then I used kinematic equations:
0 = 10 - 1.7t => t = 5.88s
s = 0 + (10 m/s)(5.88 s) + (1/2)(-1.7)(5.88)^2
That was the hypotenuse of the triangle. Using trig to solve I came up with the y axis leg = 14.3. Not the right solution.
Where did I go wrong?