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3 months ago

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A book slides across a level, carpeted floor with an initial speed of 3.25 m/s and comes to rest after 3.25 m. Calculate the coe

fficient of kinetic friction ????k between the book and the carpet. Assume the only forces acting on the book are friction, weight, and the normal force.

1 answer:

Maru [3.3K]3 months ago

8 0

Answer:

The coefficient of kinetic friction calculates to 0.165.

Explanation:

Given that,

Initial speed = 3.25 m/s

Distance traveled = 3.25 m

We must figure out the acceleration

Applying the third equation of motion

$v^2=u^2+2as$

$a=\dfrac{v^2-u^2}{2s}$

Where a = acceleration

u = starting velocity

v = ending velocity

s = distance

Insert the values into the formula

$a=\dfrac{-(3.25)^2}{2\times3.25}$

$a=-1.625\ m/s^2$

The negative sign indicates deceleration

Next, we calculate the coefficient of kinetic friction.

Using the frictional force

$F_{k} =k mg$ ....(I)

Where k = kinetic friction coefficient

m = mass

g = gravitational acceleration

Utilizing Newton's second law

$F = ma$ ....(II)

Setting equations (I) and (II) equal to each other

$k mg=ma$

$k=\dfrac{a}{g}$

$k=\dfrac{1.625}{9.8}$

$k=0.165$

Thus, the coefficient of kinetic friction is determined to be 0.165.

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