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## Homework Statement

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.

## Homework Equations

See below.

## 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

= 28.59

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?

Thank you