Projectile with friction on an inclined plane

[kraaaaamos]

Projectile motion with friction on an inclined plane -- PLEASE HELP!

Homework Statement

A 2.0 kg wood block is launched up a wooden ramp that is inclined at a 35* angle. The block’s initial speed is 10m/s. (Given that μk = 0.20)

a. What vertical height does the block reach above its starting point?
b. What speed does it have when it slides back down to its starting point?

Homework Equations

The Attempt at a Solution

FOR PART A:

I know that initial velocity (y component) is Vi = Vi sin theta
= 10 sin 35
= 5.73m/s
I know we use the equation:
Vf^2 = Vi^2 + 2a (deltaY)

So we need the value of a . . .

Fnet = ma, but we need to knwo the value of Fnet
Fnet = Flaunch - Fk?
We know that Fk = coeff (m)(g)
= 0.2 (2)(-9.8)
= -3.92N
Do we need to find teh value of the force of the launch?

I assumed that the value of the launch can be divided into the x-component and the y-component

vi(x) = Vicos theta
= 10 cos 35
= 8.19 m/s
vi(y) = Vi sin theta
= 10 sin 35
= 5.73 m/s

and if we plug in those values to get their overall magnitude

sqrt ( vi(y)^2 + vi(x)^2 )
sqrt [ (8.19)^2 + (5.73)^2 ]
sqrt [ 67.1 + 32.9]
sqrt (100)
= 10

Flaunch = 10 ( mass)
= 10 (2)
= 20N

Fnet = 20 - 3.92
= 16.08

Fnet = ma
16.08 = (2)a
a = 8.04 m/s2

Vf(y)^2 = Vi(y)^2 + 2(ay)(deltaY)
0^2 = 5.73^2 + 2 (-8.04)(deltay)
-32.8 = -16.08 (deltaY)
deltaY = 2.04m <<< FINAL ANSWER...

is that corect? Everything from the point where I suggested that Fnet = Flaunch - Fk . . . was something I deduced on my own. So I have no idea if it even makes sense.

 

[rl.bhat]

Instead of taking velocity component, take the component of weight along the inclined plane. When the block is going up Fnet = component of weight + Fk. If you want to launch the block up the inclined plane you have to do work against this force. Work done = Fnet X distance moved along the inclined plane. From that you can find the height. When the block is moving down, Fnet = Component of weight -Fk.
While going up work done = loss of KE
While going down work done = gain in KE

 

[catkin]

It is simpler to resolve velocities and forces parallel and perpendicular to the ramp rather than horizontal and vertical.

Reason: the block moves parallel to the ramp; the normal force (required to calculate the frictional force) is perpendicular to the ramp.

Using this approach I get this expression for the answer to a
$$\frac{10^{2}tan35}{2g(sin35 + 0.2cos35}$$