समस्या: 4 किग्रा द्रव्यमान का एक गुटका क्षैतिज सतह पर टिका हुआ है। यदि गुटके पर 25 N का बल लगाया जाए तो वह हिलता नहीं है। लेकिन, जैसे ही यह बल इस बिंदु से आगे बढ़ता है, ब्लॉक हिलना शुरू कर देता है। ब्लॉक और सतह के बीच घर्षण का गुणांक ज्ञात कीजिए।
Problem]: A block of mass 4 kg is resting on a horizontal surface. If a force of 25 N is applied to the block, it does not move. But, as this force is increased beyond this point the block starts moving. Find the coefficient of friction between the block and the surface.Original Question set: friction numerical class 11 set 1 (in Hindi & English)
Solution:
To find the coefficient of friction between the block and the surface, we need to consider the maximum friction force that can be applied before the block starts moving. At the point where the block is on the verge of moving, the applied force is equal to the maximum friction force.
Given:
Mass of the block (m) = 4 kg
Applied force when the block is on the verge of moving (F_applied) = 25 N
Calculations:
Maximum friction force (F_max_friction) = F_applied
Now, let’s calculate:
F_max_friction = 25 N
Again, as per friction formula, F_max_friction = μ m g = μ x 4 x 9.8
Hence, μ x 4 x 9.8 = 25
=> μ = 25/(4 x 9.8) = 0.63
So, the coefficient of friction between the block and the surface is 0.63
Also See for other such Numericals on friction: friction numerical class 11 set 1 (in Hindi & English)