Projectile with friction on an inclined planeby kraaaaamos Tags: friction, inclined, plane, projectile 

#1
Nov807, 11:01 PM

P: 20

1. The problem statement, all variables and given/known data
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? 2. Relevant equations 3. 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 xcomponent and the ycomponent 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. 



#2
Nov907, 03:55 AM

HW Helper
P: 4,442

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 



#3
Nov907, 05:04 AM

P: 218

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 [tex]\frac{10^{2}tan35}{2g(sin35 + 0.2cos35}[/tex] 


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