Motion on the Inclined Plane?

A box is sliding up an incline that makes an angle of 20.0° with respect to the horizontal. The coefficient of kinetic friction between the box and the surface of the incline is 0.191. The initial speed of the box at the bottom of the incline is 4.40 m/s. How far does the box travel along the incline before coming to rest?

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  • Amy
    Lv 7
    2 months ago

    It's useful in these problems to tilt your xy plane. Draw the x-axis parallel to the surface instead of horizontal. The y-axis is perpendicular to that. Now the motion of the object only has an x component, so everything in the y direction needs to cancel out just like on a flat surface.

    Which forces now have x components? Which forces have y components?Calculate the x- and y- components of the box's weight. What does that make the magnitude of friction? What is the net force?

    Finally, remember that work done equals the change in kinetic energy. Over how much distance does the net force act before KE is reduced to zero?

  • 2 months ago

    first get net force on box

    force of weight normal to ramp is mg cos 20

    force of weight parallel to ramp is mg sin 20

    force of friction is 0.191(mg cos 20)

    net force = mg sin 20 + 0.191(mg cos 20) pointed down the ramp

    F = mg(0.342 – 0.179) = mg(0.163)

    then use that to get acceleration

    a = F/m = g(0.163) = 1.59 m/s²

    then use motion eq to get distance.

    v² = v₀² + 2ad

    0 = 4.4² + 2(–1.59)d

    d = 6.09 m

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