Thursday
March 23, 2017

Post a New Question

Posted by on Sunday, February 3, 2013 at 12:54am.

A wooden block of mass m = 9 kg starts from rest on an inclined plane sloped at an angle θ from the horizontal. The block is originally located 5m from the bottom of the plane.

If the block, undergoing constant acceleration down the ramp, slides to the bottom in t = 2 s, and θ = 30º, what is the magnitude of the kinetic frictional force on the block?
ƒk =

  • Physics - , Tuesday, February 5, 2013 at 11:22am

    Wb = m*g = 9kg * 9.8N/kg = 88.2 N = Wt
    of the block.

    Fb = 88.2 N. @ 30o = Force of block.
    Fp = 88.2*sin30 = 44.1 N. = Force parallel to incline.
    Fv = 88.2*cos30 = 76.4 N. = Force perpendicular to the incline.

    d = 0.5a*t^2 = 5 m.
    0.5a*2^2 = 5
    2a = 5
    a = 2.5 m/s^2.

    Fp - Fk = m*a
    44.1 - Fk = 9*2.5 = 22.5
    Fk = 44.1 - 22.5 = 21.6 N.










































    fp b=

Answer This Question

First Name:
School Subject:
Answer:

Related Questions

More Related Questions

Post a New Question