At the end of a factory production line, boxes start from rest and slide down a 30 degree ramp 5.2 m long.?

If the slide is to take no more than 4.3 s , what's the maximum allowed frictional coefficient?

2 Answers

  • oubaas
    Lv 7
    4 months ago
    Favorite Answer

    acceleration = 2d/t^2 = 10.4/4.3^2 = 0.562

    accel. a = (motive force-friction force)/mass

    0.562 = m*g*(sin30-cos30*μ)/m

    mass m cross

    0.562 = 4.90-8.49*μ

    μ = (4.90-0.562)/8.49 = 0.511

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  • 4 months ago

    First, we find the maximum acceleration. Since the box starts from rest and accelerates down the plane, we can use

    s = 1/2 a t^2

    to solve for acceleration, since you know the distance, s, is 5.2m and t is 4.3 s Solve for a, we will use it below.

    the forces acting along the ramp are the component of gravity down the ramp, given by m g sin (theta), and the frictional force up the ramp which is given by u m g cos(theta) where u is the coeff of friction.

    Your book should derive these equations fully.

    Newton's second law tells you that ma equals the sum of forces acting on the box, so we have

    m a = m g sin(theta) - u m g cos(theta)

    where we have chosen down the ramp as the positive direction

    this tells us that

    a = g( sin (theta) - u cos(theta))

    now, use the value of a you computed above with theta = 30 degrees and g = 9.8 m/s/s and solve for u

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