One block of 2.0 kg mass is placed on top of another block of 3.0 kg mass. The coefficient of static friction between the two blocks is 0.2. The bottom block is pulled with a horizontal force F such that both the blocks move together without slipping. Taking acceleration due to gravity as 10 m/s2, the maximum value of the frictional force is :
- (a)50 N
- (b)30 N
- (c)4 N
- (d)10 N
Correct — D, 10 N (as marked in the official key). The friction between the blocks can supply at most f = mu x m(top) x g = 0.2 x 2.0 x 10 = 4 N to the upper block, giving a maximum shared acceleration a = f/m(top) = 4/2 = 2 m/s2. The largest horizontal force that can be applied to the bottom block while both still move together without slipping is therefore F = (2.0 + 3.0) x 2 = 10 N. (Read literally, the peak friction between the blocks is the 4 N in option (c); the key's 10 N is this maximum driving force F.)
- (a)50 N — 50 N is the total weight of both blocks (5 kg x 10 m/s2) — the normal force on the ground, not the driving force or the inter-block friction.
- (b)30 N — Does not correspond to any correct quantity here; it is neither the limiting friction (4 N) nor the maximum driving force (10 N).
- (c)4 N — This is the maximum FRICTION between the two blocks (mu x m(top) x g). It is the correct value if the phrase is read literally, but the official key marks the maximum driving force F (10 N) instead.
When a lower block is pulled and an upper block rides on it, only friction accelerates the upper block. Static friction can grow up to a limit f(max) = mu x N, where N is the weight of the upper block. That limiting friction fixes the greatest common acceleration; beyond it the top block slips. The maximum force that can be applied to the system while they still move together is F = (m1 + m2) x a(max).
Two numbers appear: the limiting friction between the blocks (4 N) and the maximum applied force F (10 N). The official key reports 10 N. The safe way to learn this item is to compute both and see how the friction limit (4 N -> a = 2 m/s2) leads to F = 10 N.
- Limiting static friction between the blocks: f = mu x m(top) x g = 0.2 x 2 x 10 = 4 N.
- Maximum common acceleration: a = f/m(top) = 2 m/s2.
- Maximum driving force for no slipping: F = (m1 + m2) a = 5 x 2 = 10 N (official answer).
- Static friction adjusts up to its limit; beyond it the upper block slides.
- Read carefully: the maximum friction between the blocks is 4 N (option c); the official key's 10 N is the maximum applied force F for no slipping.
- Using the total weight (50 N) or an unrelated value.
Asked as the limiting friction / maximum force for two stacked blocks to move together without slipping.
No directly related past PYQ was found.
- practice — not a real PYQ
For a 2 kg block on a lower block, mu = 0.2, g = 10 m/s2, the maximum friction the surface can provide to the upper block is
- (a)4 N
- (b)10 N
- (c)20 N
- (d)40 N
Answer(a) 4 N (mu x m x g).
- practice — not a real PYQ
If the limiting friction gives a maximum shared acceleration of 2 m/s2, the maximum force to move a 5 kg two-block system together is
- (a)5 N
- (b)10 N
- (c)20 N
- (d)2 N
Answer(b) 10 N (F = ma = 5 x 2).