समस्या: 0.5 किग्रा द्रव्यमान वाली एक गेंद क्षैतिज सतह पर 2 मीटर/सेकेंड के वेग से घूम रही है। रोलिंग घर्षण का गुणांक 0.1 है। गेंद की मंदी ज्ञात कीजिए।
Problem]: A ball with a mass of 0.5 kg is rolling on a horizontal surface with a velocity of 2 m/s. The coefficient of rolling friction is 0.1. Find the deceleration of the ball.Original Question set: friction numerical class 11 set 1 (in Hindi & English)
Solution:
To calculate the deceleration of the rolling ball, we need to consider the forces acting on it. These forces include the force of rolling friction, which opposes the motion of the ball. We will apply Newton’s second law to determine the deceleration.
Given:
Mass of the ball (m) = 0.5 kg
Velocity of the ball (v) = 2 m/s
Coefficient of rolling friction (μ) = 0.1
Calculations:
Force of rolling friction (F_rolling_friction) = μ * m * g = 0.1 * 0.5 kg * 9.8 m/s² ≈ 0.49 N
Deceleration (a) = F_rolling_friction / m = 0.49 N / 0.5 kg ≈ 0.98 m/s²
The deceleration of the rolling ball is approximately 0.98 m/s².
Also See for other such Numericals on friction: friction numerical class 11 set 1 (in Hindi & English)