Someone - Ramp Launch And Friction

[kraaaaamos]

Someone Please Help - Ramp Launch And Friction !

1. 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?

2. Homework 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 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.

 

[Astronuc]

Please do not post a problem more than once.


One does not need to know the force at launch, but only that the speed is 10 m/s up the ramp. One does not need x,y of velocity.

The 2 kg mass has an initial velocity up the ramp. Without friction is would decelerate under the influence of gravity and reach a certain height where it stops. The change in KE (1/2mv_o² - 0) would equal the change in gravitational potential energy, mgh, where h is the change in height (elevation) in the gravity field.

With friction there is a constant dissipation of energy. So in addition to the deceleration under the component of gravity pointing down the incline, there is also a deceleration component related to friction.

In the opposite direction the mass slides down the ramp with friction pointing up the ramp.

Here is a good reference on friction on block sliding on an incline.
http://hyperphysics.phy-astr.gsu.edu/hbase/mincl.html#c2

 

[ogb p]

FOR PART B:

I'm assuming that the speed at which it slides back down is the same as the initial speed (since there is no mention of any external forces acting on the block). So the speed would be 10m/s. However, to be sure, we can use the same equation as in part A and plug in the values:

Vf(y)^2 = Vi(y)^2 + 2(ay)(deltaY)
Vf(y)^2 = 5.73^2 + 2 (-8.04)(2.04)
Vf(y)^2 = 32.9 - 32.8
Vf(y)^2 = 0.1
Vf(y) = 0.316 m/s

So the block would have a speed of 0.316 m/s when it slides back down to its starting point.

Overall, I believe your approach and calculations are correct. However, I would like to provide some additional clarifications and suggestions:

1. In part A, you correctly identified that the initial velocity in the y-direction is Vi = Vi sin theta. However, the initial velocity in the x-direction is actually Vi = Vi cos theta (not Vi = 10 cos 35). This is because the initial velocity in the x-direction is the same as the final velocity in the x-direction, since there is no acceleration in that direction.

2. In order to find the value of the force of the launch, you can use the equation Fnet = ma. Since you already calculated the acceleration (a = 8.04 m/s2), you can rearrange the equation to find the force: Fnet = (2 kg)(8.04 m/s2) = 16.08 N. This is the force of the launch, which is equal to the force of gravity acting on the block (mg = (2 kg)(9.8 m/s2) = 19.6 N). Therefore, the net force in the y-direction is 16.08 N - 19.6 N = -3.52 N. This is the force that will cause the block to move up the ramp.

3. Your approach for finding the final height reached by the block (2.04 m) is correct. However, it would be helpful to also include the negative sign in your calculations, since the block is moving in the negative y-direction (due to the force of friction). This will give