# 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?

Relevance

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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• 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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