91Ó°ÊÓ

$\mathrm{θ}={\text{25}}^{\text{o}}$

$\mathrm{a}=0.80{\text{″¾/²õ}}^{\text{2}}$

$\mathrm{W}=40\text{ N}$

The problem is based on Newton’s second law of motion. It states that the rate of change in the momentum of a body is directly proportional to the magnitude and direction of the force acting on the body which implies the acceleration of the body with force and the mass of the body.

Formula:

The force according to Newton’s second law, ${\mathrm{F}}_{\mathrm{net}}=\mathrm{ma}$ …(¾±)

The weight of the body, $\mathrm{W}=\mathrm{mg}$ …(¾±¾±)

The coefficient of kinetic friction, ${\mathrm{μ}}_{\mathrm{K}}={\mathrm{f}}_{\mathrm{k}}/\mathrm{N}$ …(¾±¾±¾±)

The mass of the black is given by,

$\begin{array}{c}\mathrm{m}=\frac{\mathrm{W}}{\mathrm{g}}\\ =4\text{ k²µ}\end{array}$

From the free body diagram, the kinetic friction between the block and the plane can be calculated using equation (i) as follows:

$\begin{array}{c}{\mathrm{F}}_{\mathrm{net}}=\mathrm{ma}\\ ={\mathrm{f}}_{\mathrm{k}}−\mathrm{°Â²õ¾\pm ²Ôθ}\\ {\mathrm{f}}_{\mathrm{k}}=\mathrm{ma}+\mathrm{°Â²õ¾\pm ²Ôθ}\\ =\left(4\text{kg}\right)\left(0.80{\text{m/s}}^2\right)+\left(40\text{N}\right)\sin \left({25}^0\right)\\ =20.10\text{N}\\ ≈20\text{N}\end{array}$

Now, applying the force equations to the y-axis of the free body diagram, we can get the normal force on it as follows:

$\begin{array}{c}{\stackrel{→}{\mathrm{F}}}_{\mathrm{y}}=0\\ \mathrm{N}=\mathrm{°Â³¦´Ç²õθ}\\ =\left(40\text{N}\right)\cos \left({25}^0\right)\\ =36.25\text{N}\end{array}$

Thus, the coefficient of friction can be given as follows:

$\begin{array}{c}{\mathrm{μ}}_{\mathrm{K}}=20\text{N}/36.25\text{N}\\ =0.552\text{N}\\ =0.55\text{N}\end{array}$

Hence, the value of the coefficient of the kinetic friction is$0.55\text{N}$ .