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19 July, 00:26

A 562 N trunk is on frictionless plane inclined at 30.0 degrees from the horizontal. What is the acceleration of the trunk down the ramp?

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  1. 19 July, 02:00
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    Answer: 0m/s²

    Explanation:

    Since the forces acting along the plane are frictional force (Ff) and moving force (Fm), we will take the sum of the forces along the plane

    According newton's law of motion

    Summation of forces along the plane = mass * acceleration

    Frictional force is always acting upwards the plane since the body will always tends to slide downwards on an inclined plane and the moving acts down the plane

    Ff = nR where

    n is coefficient of friction = tan (theta)

    R is normal reaction = Wcos (theta)

    Fm = Wsin (theta)

    Substituting in the formula of newton's first law we have;

    Fm-Ff = ma

    Wsin (theta) - nR = ma

    Wsin (theta) - n (Wcos (theta)) = ma ... 1

    Given

    W = 562N, theta = 30°, n = tan30°, m = 56.2kg

    Substituting in eqn 1,

    562sin30° - tan30° (562cos30°) = 56.2a

    281 - 281 = 56.2a

    0 = 56.2a

    a = 0m/s²

    This shows that the trunk is not accelerating
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