(a) A box of 5.0 kg slides down a 30° inclined plane with an acceleration of 1.2

Question

(a) A box of 5.0 kg slides down a 30° inclined plane with an acceleration of 1.2 m/s2. Determine the coefficient of kinetic friction between the box and the inclined plane.

(b) Two objects of masses 10.0 kg and 5.0 kg are connected by a light string that passes over a frictionless pulley installed at the tip of a frictionless incline of angle 30°. The 5.0 kg object lies on the inclined plane. Assume that the 10.0 kg object is moving downward in the air. Find the acceleration of each object and the tension in the string.

Explanation / Answer

normal force N = m*g*cos30

frictional force up the plane = uk*N

component pf gravitational force down the plane = m*g*sintheta

Fnet = Fg - fk

Fnet = ma

ma = m*g*sintheta - uk*m*g*costheta

a = g*(sintheta - uk*costheta)

1.2 = 9.8*(sin30 - (uk*cos30))

uk = 0.435 <-------answer

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(b)

for hanging mass

10*9.8 - T = 10*a

T = 10*(9.8-a) ,,,,,,,,,,,,(1)

for the mass onincline

T - 5*9.8*sin30 = 5*a..........(2)

substituting 1 in 2

10*(9.8-a) - (5*9.8*sin30) = 5*a

a = 4.9 m/s^2 <<---------answer

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