Question

A student wants to determine the coefficients of static friction and kinetic friction between a box and a plank. She places the box on the plank and gradually raises one end of the plank. When the angle of inclination with the horizontal reaches$\mathbf{30}\mathbf{°}$, the box starts to slip, and it then slides 2.5 mdown the plank in 4.0 sat constant acceleration. What are

(a) the coefficient of static friction and

(b) the coefficient of kinetic friction between the box and the plank?

Short answer

$$\begin{gathered} \mathbf{a}\mathbf{)}\mathbf{}\mathbf{}\mathbf{}{\mu }_s=0.58. \\ \mathbf{b}\mathbf{)}\mathbf{}{\mu }_K=0.54. \end{gathered}$$

Step-by-step solution

Given data

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

Understanding the concept

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:

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

Draw the free body diagram

(a) Calculate the coefficient of static friction

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

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

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

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

Thus, the coefficient of static friction is 0.58.

(b) Calculate the coefficient of kinetic friction between the box and the plank

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:

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

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

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

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

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

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

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

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

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