91Ó°ÊÓ

• The angle of inclination of plank surface with horizontal$\mathrm{θ}=30°$.
• Displacement of the box down the plank surface 0.5 m.
• Time t = 4.0 s.

The problem deals with Newton’s second law of motion, which states that the acceleration of an object is dependent upon the net force acting upon the object and the mass of the object. Also, it involves the kinematic equation of motion in which the motion is described at constant acceleration.Use Newton's second law and kinematic equations of motion to determine the coefficient of static friction and coefficient of kinetic friction.

Formula:

$x=\left(v_i×t\right)+\frac{1}{2}a{t}^2$

For the condition that the box just starts to slip (static friction condition),

$$\begin{gathered} f=mg\sin \left(30°\right) \\ {\mathrm{μ}}_s{F}_n=mg\sin \left(30°\right) \\ {\mathrm{μ}}_s=\frac{mg\sin \left(30°\right)}{F_n} \end{gathered}$$

Substitute the values in the above expression, and we get,

$$\begin{gathered} {\mathrm{μ}}_s=\frac{mg\sin \left(30°\right)}{mg\cos \left(30°\right)} \\ =\frac{\sin \left(30°\right)}{\cos \left(30°\right)} \\ {\mathrm{μ}}_s=\tan \left(30°\right) \\ {\mathrm{μ}}_s=0.577 \\ ≈0.58 \end{gathered}$$

Thus, the coefficient of static friction is 0.58.

When the box is sliding down the plank with constant acceleration:

Applying Newton’s second law:

$mg\sin \left(30°\right)-f=ma$ (i)

To find the acceleration, we have to use the Kinematical equation:

$x=\left(v_i×t\right)+\frac{1}{2}a{t}^2$

Here x = 2.5 m , t = 4.0 s , v_i= 0 m/s

Substitute the values in the above expression, and we get,

$$\begin{gathered} 2.5\mathrm{m}=0+\left(0.5×a×{\left(4.0s\right)}^2\right) \\ \mathrm{a}=0.3125\mathrm{m}/{\mathrm{s}}^2 \end{gathered}$$

Using this value in equation (i), we get,

$$\begin{gathered} mg\sin \left(30°\right)-{\mathrm{μ}}_K{F}_n=ma \\ mg\sin \left(30°\right)-ma={\mathrm{μ}}_K×mg\cos \left(30°\right) \\ {\mathrm{μ}}_K=\frac{mg\sin \left(30°\right)-ma}{mg\cos \left(30°\right)} \\ {\mathrm{μ}}_K=\frac{g\sin \left(30°\right)-a}{g\cos \left(30°\right)} \end{gathered}$$

Substitute the values in the above expression, and we get,

$$\begin{gathered} {\mathrm{μ}}_k=\frac{9.8m/s^2×\sin \left(30°\right)-0.3125m/s^2}{9.8m/s^2×\cos \left(30°\right)} \\ {\mathrm{μ}}_K=0.54 \end{gathered}$$

Thus, the coefficient of kinetic friction between the block and plank is 0.54.