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March 30, 2017

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A block of mass 0.5 kg rests on a smooth inclined plane. It is then pushed by a horizontal force of 16N. When the block has displaced 5m along the inclined plane.

What would be the speed of the block if the plane is rough and a friction of 5N is acting on the block?

  • Q. incomplete and/or not clear - ,

    It is not clear to me
    1. how a block will "rest" on a smooth inclined plane, and
    2. in which direction is the horizontal force acting, towards up the plane, or down?
    3. The first paragraph seems incomplete

  • Physics - ,

    I don't know.....it's written like that in the book...... It's being pushed up....... the answer is 12.9m/s.....But I dun know how to calculate it......

  • Physics - ,

    Do you have a diagram that shows the angle of the incline and the direction of the force?

  • Physics - ,

    no....sorry I forgot to mention, the angle is 20 degree.....

  • Physics - ,

    Resolve forces along the inclined plane of inclination θ=20°:

    For the mass, m = 0.5 kg
    and the applied force, F = 16N
    component of force down the incline,
    Fm = mg*sin(θ)

    Due to force F, component up the incline,
    Ff = Fcos(θ)

    Net force up,
    Fup
    = Ff-Fm
    = Fcos(θ)-mgsin(θ)

    Acceleration up the plane
    a = F/m

    Initially at rest, u=0, and final velocity v after travelling a distance of S = 5m,
    use the relation
    v²-u² = 2aS

    Solve for v in m/s. I get 12.9 m/s.

  • Physics - ,

    a= F/m
    =16/0.5
    =32m/s
    v²=2*32*5
    v =17.9

    But I get 17.9m/s..... acceleration is equal to net force over mass or just force over mass?

    Anything wrong with my steps?

    Can I minus 17.9 by 5N which is the friction?

  • Physics - ,

    I get it now......Thanks alot.....

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