Question

A force of 400-N pushes on a 25-kg box horizontally. The box accelerates at 9 m/s? Find the coefficient of kinetic friction between the box and floor.

Answer 1

Answer:

0.69

Explanation:

Using the Newtons law of motion;

$\sum Fx = ma_x\\Fm - Ff = ma_x$

Fm is the moving force = 400N

Ff is the frictional force = μR

μ is the coefficient of kinetic friction

R is the reaction = mg

m is the mass

a is the acceleration

The equation becomes;

$Fm - \mu R = ma_x\\Fm - \mu mg = ma_x\\400- \mu (25)(9.8) = 25(9)\\400 - 254.8 \mu = 225\\- 254.8 \mu = 225 - 400\\- 254.8 \mu = -175\\ \mu = \frac{-175}{- 254.8} \\\mu = 0.69$

Hence the coefficient of kinetic friction between the box and floor is 0.69

Answer 2

The coefficient of kinetic friction between the box and the floor is 0.714.

Friction: This can be defined as the force that tends to oppose two surfaces in motion.

The question above can be solved using the formula

F-ma = mgμ................. Equation 1

Where F = Force applied to push the box, m = mass of the box, a = acceleration of the box, μ = coefficient of kinetic friction, g = acceleration due to gravity.

make μ the subject of the equation

μ = (F-ma)/mg................ Equation 2

From the question,

Given: F = 400 N, m = 25 kg, a = 9 m/s²,

Constant: g = 9.8 m/s²

Substitute these values into equation 2

μ = [400-(25×9)]/(25×9.8)

μ = (400-225)/245

μ = 175/245

μ = 0.714

Hence, the coefficient of kinetic friction between the box and the floor is 0.714.

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