समस्या: 5 किलो द्रव्यमान का एक ब्लॉक क्षैतिज सतह पर रखा गया है। ब्लॉक पर 20 N का बल लगाया जाता है। ब्लॉक और सतह के बीच घर्षण का गुणांक 0.3 है। ब्लॉक के त्वरण और उस पर लगने वाले घर्षण बल की गणना करें।
Problem 1] A block of mass 5 kg is placed on a horizontal surface. A force of 20 N is applied to the block. The coefficient of friction between the block and the surface is 0.3. Calculate the acceleration of the block and the force of friction acting on it.Original Question set: friction numerical class 11 set 1 (in Hindi & English)
Solution:
To find the acceleration of the block, we will use Newton’s second law of motion: F = ma, where F is the net force, m is the mass of the block, and a is the acceleration. The net force can be calculated as the difference between the applied force and the force of friction.
Given:
Mass of the block (m) = 5 kg
Applied force (F_applied) = 20 N
Coefficient of friction (μ) = 0.3
Calculations:
Force of friction (F_friction) = μ * m * g
- F_friction = 0.3 * 5 kg * 9.8 m/s² ≈ 14.7 N
Net force (F_net) = F_applied – F_friction
- F_net = 20 N – 14.7 N ≈ 5.3 N
Acceleration (a) = F_net / m
Acceleration (a) = 5.3 N / 5 kg ≈ 1.06 m/s²
Thus, the acceleration of the block is approximately 1.06 m/s², and the force of friction acting on it is approximately 14.7 N.
Also See for other such Numericals on friction: friction numerical class 11 set 1 (in Hindi & English)