Block on an Inclined Plane with Friction

Problem Statement

A block of mass 5 kg is placed on a rough inclined plane that makes an angle of 30 degrees with the horizontal. The coefficient of kinetic friction between the block and the plane is 0.2.

1. Find the acceleration of the block as it slides down the plane.
2. If the block starts from rest and slides for 3 s, find the distance it travels along the plane.

(Take g = 9.8 m/s^2.)

Given / Required

Given:

• m = 5 kg
• theta = 30 degrees
• mu_k = 0.2
• g = 9.8 m/s^2

Required:

1. Acceleration a of the block along the incline.
2. Distance s traveled in t = 3 s.

Hint

Resolve the weight into components parallel and perpendicular to the incline. Friction acts up the plane because the block slides downward.

Step-by-Step Solution

Step 1 – Resolve the Weight

\[ W = mg = 5 \times 9.8 = 49,\text{N} \]

\[ W_{\parallel} = mg \sin 30^\circ = 24.5,\text{N} \]

\[ W_{\perp} = mg \cos 30^\circ \approx 42.4,\text{N} \]

Step 2 – Normal Reaction and Friction

\[ N \approx 42.4,\text{N} \]

\[ f_k = \mu_k N = 0.2 \times 42.4 \approx 8.48,\text{N} \]

Step 3 – Net Force Along the Plane

\[ F_{\text{net}} = W_{\parallel} - f_k \approx 24.5 - 8.48 = 16.02,\text{N} \]

Step 4 – Acceleration

\[ a = \frac{F_{\text{net}}}{m} \approx \frac{16.02}{5} \approx 3.2,\text{m/s}^2 \]

Step 5 – Distance in 3 Seconds

\[ s = ut + \frac{1}{2} a t^2 = 0 + 0.5 \times 3.2 \times 9 = 14.4,\text{m} \]

Final Answer

• Acceleration: a approx 3.2 m/s^2.
• Distance in 3 s: s approx 14.4 m.

Extension / Variation

• Repeat with a smooth plane (no friction).
• Consider the block pushed up the incline initially.

Key Concept Recap

• Resolve weight into components.
• Friction opposes motion along the surface.
• Apply F_net = m a along the direction of motion.