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Homework Help: Someone Please Help - Ramp Launch And Friction !

  1. Nov 9, 2007 #1
    Someone Please Help - Ramp Launch And Friction !!!!!!!!!!!!!!!!!!!!!

    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


    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.
  2. jcsd
  3. Nov 9, 2007 #2


    User Avatar
    Staff Emeritus
    Science Advisor

    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 (1/2mvo2 - 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.
    Last edited: Nov 9, 2007
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