Sebuah benda yang beratnya w meluncur ke bawah dengan kecepatan tetap pada suatu bidang miring yang kasar. Bidang miring tersebut membentuk sudut 30° dengan horizontal. Berapa koefisien gesekan antara benda dan bidang miring tersebut?

Here's the solution to question 3:

The problem describes an object of weight $w$ sliding down a rough inclined plane at a constant velocity. The inclined plane makes an angle of $30^\circ$ with the horizontal. We need to find the coefficient of friction between the object and the inclined plane.

Step 1: Identify the forces acting on the object. Since the object is sliding down at a constant velocity, the net force on it is zero. The forces acting on the object are: • Weight ($w$) acting vertically downwards. • Normal force ($N$) acting perpendicular to the inclined plane. • Frictional force ($f_k$) acting parallel to the inclined plane, opposing the motion (up the incline).

Step 2: Resolve the weight into components parallel and perpendicular to the inclined plane. The component of weight parallel to the incline is $w \sin \theta$. The component of weight perpendicular to the incline is $w \cos \theta$. Here, $\theta = 30^\circ$.

Step 3: Apply Newton's First Law in the direction perpendicular to the inclined plane. Since there is no acceleration perpendicular to the plane, the normal force balances the perpendicular component of the weight. $N = w \cos \theta$

Step 4: Apply Newton's First Law in the direction parallel to the inclined plane. Since the object moves at a constant velocity, the net force parallel to the plane is zero. The component of weight pulling the object down the incline is balanced by the kinetic friction force acting up the incline. $w \sin \theta = f_k$ We know that the kinetic friction force is given by $f_k = \mu_k N$, where $\mu_k$ is the coefficient of kinetic friction. Substituting $f_k$: $w \sin \theta = \mu_k N$

Step 5: Substitute the expression for $N$ from Step 3 into the equation from Step 4. $w \sin \theta = \mu_k (w \cos \theta)$

Step 6: Solve for the coefficient of kinetic friction, $\mu_k$. Divide both sides by $w \cos \theta$: $\mu_k = \frac{w \sin \theta}{w \cos \theta}$ $\mu_k = \frac{\sin \theta}{\cos \theta}$ $\mu_k = \tan \theta$ Substitute the given angle $\theta = 30^\circ$: $\mu_k = \tan 30^\circ$ $\mu_k = \frac{1}{\sqrt{3}}$ $\mu_k = \frac{\sqrt{3}}{3}$

The coefficient of friction between the object and the inclined plane is $\frac{\sqrt{3}}{3}$.

$\boxed{\mu_k = \frac{\sqrt{3}}{3}}$

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