समस्या: 2 किग्रा द्रव्यमान का एक लकड़ी का गुटका 30 डिग्री के कोण वाले झुके हुए तल पर रखा गया है। ब्लॉक और तल के बीच घर्षण का गुणांक 0.2 है।
1)जब कोई अतिरिक्त बाहरी बल नहीं लगाया जाता है तो ब्लॉक का त्वरण और उस पर लगने वाले घर्षण बल का पता लगाएं।
2) अब. यदि ब्लॉक को ऊपर की ओर धकेलने के लिए झुके हुए तल के समानांतर 15 N का बल लगाया जाता है, तो ब्लॉक के त्वरण का परिमाण और दिशा क्या है?
Problem] A wooden block of mass 2 kg is placed on an inclined plane with an angle of 30 degrees. The coefficient of friction between the block and the plane is 0.2.
1) Find the acceleration of the block and the force of friction acting on it when no additional external force is applied.
2) Now. if a force of 15 N is applied parallel to the inclined plane to push the block upward, then what is the magnitude and direction of the acceleration of the block?Original Question set: friction numerical class 11 set 1 (in Hindi & English)
Solution:
To determine the acceleration of the block on the inclined plane, we must first resolve the forces acting on it. These forces include the force of gravity, the normal force, the applied force, and the force of friction. Once we have resolved these forces, we can apply Newton’s second law to find the acceleration.
Given:
Mass of the block (m) = 2 kg
The angle of the inclined plane (θ) = 30 degrees
Applied force (F_applied) = 15 N
Coefficient of friction (μ) = 0.2
Calculations:
Force of gravity (F_gravity) = m * g * sin(θ) = 2 kg * 9.8 m/s² * sin(30°) ≈ 9.8 N
Normal force (F_normal) = m * g * cos(θ) = 2 kg * 9.8 m/s² * cos(30°) ≈ 16.96 N
Force of friction (F_friction) = μ * F_normal = 0.2 * 16.96 N ≈ 3.392 N
Net force (F_net) = F_applied – F_friction= 15 N – 3.392 N ≈ 11.608 N
Acceleration (a) = F_net / m
Acceleration (a) = 11.608 N / 2 kg ≈ 5.804 m/s²
Thus, the acceleration of the block on the inclined plane is approximately 5.804 m/s², and the force of friction acting on it is approximately 3.392 N.
Also See for other such Numericals on friction: friction numerical class 11 set 1 (in Hindi & English)