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Answer:
<em>2.78m/s²</em>
Explanation:
Complete question:
<em>A box is placed on a 30° frictionless incline. What is the acceleration of the box as it slides down the incline when the co-efficient of friction is 0.25?</em>
According to Newton's second law of motion:
Where:
is the coefficient of friction
g is the acceleration due to gravity
Fm is the moving force acting on the body
Ff is the frictional force
m is the mass of the box
a is the acceleration'
Given
Required
acceleration of the box
Substitute the given parameters into the resulting expression above:
Recall that:
9.8sin30 - 0.25(9.8)cos30 = ax
9.8(0.5) - 0.25(9.8)(0.866) = ax
4.9 - 2.1217 = ax
ax = 2.78m/s²
<em>Hence the acceleration of the box as it slides down the incline is 2.78m/s²</em>
Answer: N = Mgcos(theta)
Therefore, the Normal reaction force is equal to Mgcos(theta)
Explanation:
See attached for a sketch.
From the attachment.
.
N = normal reaction force on block
W = weight of the block
theta = angle of the inclined plane to the horizontal
From the sketch, we can see that
N is equal in magnitude but opposite direction to Wy
N = Wy
And
Wy = Wcos(theta)
Wx = Wsin(theta)
Then,
N = Wy = Wcos(theta)
But W = mass × acceleration due to gravity = mg
N = Mgcos(theta)
Therefore, the Normal reaction force is equal to Mgcos(theta)
