Tiêu đề 1. Friction (S.C.Q. E).pdf ✅

PHYSICS Q. 1 Two blocks of masses 10 kg and 2 kg are connected by an ideal spring of spring constant 1000 N/m and the system is placed on a horizontal surface as shown. 10kg μ=0.5 2kg μ=0 The coefficient of friction between 10 kg block and surface is 0.5 but friction is assumed to be absent between 2 kg and surface. Initially blocks are at rest and spring is unstreached then 2 kg block is displaced by 1 cm to elongate the spring then released. Then the graph representing magnitude of frictional force on 10 kg and time t. (time t is measured from that instant when 2kg is released to move) (A) t f (B) t f (C) t f (D) None of these [B] Q. 2 The block of mass m is placed on a rough horizontal floor and it is pulled by a ideal string as shown by a constant force F. As the block moves towards right the friction force on block- m F (A) remains constant (B) increases (C) decreases (D) can be said [B] `Q. 3 Two blocks each of mass 20 kg are connected by an ideal string and this system is kept on rough horizontal surface as shown. Initially the string is just tight then a horizontal force F = 120 N is applied on one block as shown. 20kgd 20kgd F=120N μ = 0.5 μ = 0.5 If friction coefficient at every contact is μ = 0.5 then which of the following represents the correct free body diagram. (A) N=200N 20kgd 20kgd N= 200N T=50N F= 120N 200N F2 = 70N 200N F1 = 50 N T=50N (B) N=200N 20kgd 20kgd N= 200N T=20N F= 120N 200N F2 = 100N 200N F1 = 20 N T=20N (C) N=200N 20kgd 20kgd N= 200N T=60N F= 120N 200N F2 = 60N 200N F1 = 60 N T=60N (D) All of the above [B] Q. 4 A block of mass m is moving on a rough horizontal surface where coefficient of friction between block and surface is μ = 0.5. At t = 0 block is moving with a velocity  v then a horizontal force F = 2t is applied on it in the direction of velocity. Here t represent time then the speed of block after the application of force (A) decreases (B) increases (C) first decreases then increases (D) first increases then decreases [C]
Q. 5 A block of mass m = 1kg moving on horizontal surface with speed u = 2m/s enters a rough horizontal patch ranging from x = 0.10 m to x = 2.00m. If the retarding force fr on the block in this range is inversely proportional to x over this range i.e. fr = x k 0.10 < x < 2.00 = 0 for x < 0.10 and x > 2.00 If k = 0.5 J then the speed of this block as it crosses the patch is (use n 20 = 3) (A) 2.65 m/s (B) 1 m/s (C) 1.5 m./s (D) 2 m/s Sol. [B] Use W = K where W =  2.0 0.1 f.dx Q.6 A block of mass 70 kg is kept on a rough horizontal surface and coefficient of static friction between block and surface is 0.4. A man is trying to pull the block by applying a horizontal force. If the surface exerts a force F on block then – (A) F must be 700 N (B) F must be 280 N (C) 700 N  F  754 N (D) F may be greater than 754 N [C] Q.7 A block of mass m1 = 20 kg is kept on horizontal surface where coefficient of friction between block and surface is μ = 0.75. This block is connected by an ideal string with another block of mass m2. The string is passing over an ideal fixed pulley and the string between pulley and m2 is vertical as shown in the figure. Then the minimum value of mass m2 which can make m1 just move on horizontal surface is (position of m1 on surface is not fixed you can adjust it to get minimum value of m2 and assume g = 10 m/s2 ). μ ///////////////////////////////////////\\\\\\\\\ m1 m2 \\\\\\\\\\\\ (A) 12 kg (B) 14 kg (C) 16 kg (D) 20 kg [A] `Q. 8 A block of mass 1 kg starts slipping on a circular track of radius 2m and it is observed that when  = 60o its speed is 4 m/s as shown in figure. Assuming size of block to be negligible and coefficient of friction between block and track is 0.5 frictional force on block when  = 60o is (g = 10 m/s2 ) –  O r =2m v (A) 5 N (B) 2.5 N (C) 6.5 N (D) 4 N [C] Q. 9 A particle moves on a rough horizontal ground with some initial velocity say v0. If 3/4th of its kinetic energy is lost in friction in time t0. Then coefficient of friction between the particle and the ground is - (A) 0 0 2gt v (B) 0 0 4gt v (C) 0 0 4gt 3v (D) 0 0 gt v [A] Sol. 4 3 th energy is lost i.e., 4 1 th kinetic energy is left. Hence, its velocity becomes 2 v0 under a retardation of g in time t0.  2 v0 = v0 – g t0 or g t0 = 2 v0 or  = 0 0 2g t v Q.10 A block A of mass 2 kg rests on another block B of mass 8 kg which rests on a horizontal floor. The coefficient of friction between A and B is 0.2 while between B and floor is 0.5. When a horizontal force 25 N is applied on B, the frictional force acting between A and B is - (A) 0 N (B) 3.9 N (C) 5 N (D) 49 N [A] Sol. Limiting friction on block B = 0.5 × (2 + 8) × 10 = 50 N
 block B will not move. Q.11 A block starts moving up a fixed inclined plane of inclination 60° with a velocity of 20 m/s and stops after 2 sec. The approximate value of coefficient of friction is (g = 10 m/s2 ) (A) 3 (B) 3.3 (C) 0.27 (D) 0.33 [C] Sol. Retardation = g (sin  +  cos ) = 5 ( 3 ) Now v = u – at  a = t u as v = 0  5( 3 ) = 10   = 0.27 Q.12 A rod AB of mass 10 kg tied with a string at C such that AC = BC and rod remains in equilibrium then friction force by horizontal surface will be – C B A (A) 50 N towards right (B) 50 N towards left (C) 100 N (D) None of these [A] Sol. A f 45o N mg B T C T = f ....(1) N = mg ....(2) taking torque about A T × 2 L = mg × 2 L / 2  T = 2 mg = 50 N Q.13 In the given figure the wedge is fixed, pulley is frictionless and string is light. Surface AB is frictionless whereas AC is rough. If the block of mass 3m slides down with constant velocity, then the coefficient of friction between surface AC and the block is – 3m m 45° 45° A B C (A) 1/3 (B) 2/3 (C) 1/2 (D) 4/3 [B] Q.14 A block of mass m is pulled by a constant power P placed on a rough horizontal plane. The friction co-efficient between the block and the surface is Maximum velocity of the block will be – (A) mg P (B) P mg (C) mgP (D) mg P  [D] Sol. P = F.v = constant When F = mg, net force on block becomes zero. i.e. it has maximum or terminal velocity  P = (mg)vmax or vmax = mg P  Q.15 A block of mass m is moving on a rough horizontal surface where coefficient of friction between block and surface is μ = 0.5. At t = 0 block is moving with a velocity  v then a horizontal force F = 2t is applied on it in the direction of velocity. Here t represent time then the speed of block after the application of force- (A) decreases (B) increases (C) first decreases then increases (D) first increases then decreases [C] Sol. FNet = F – fk = 2t – mg Q. 16 Two blocks each of mass 20 kg are connected by an ideal string and this system is kept on rough horizontal surface as shown. Initially the string is just tight then a horizontal force F = 120 N is applied on one block as shown.
20kgd 20kgd F=120N μ = 0.5 μ = 0.5 If friction coefficient at every contact is μ = 0.5 then which of the following represents the correct free body diagram. (A) N=200N 20kgd 20kgd N= 200N T=50N F= 120N 200N F2 = 70N 200N F1 = 50 N T=50N (B) N=200N 20kgd 20kgd N= 200N T=20N F= 120N 200N F2 = 100N 200N F1 = 20 N T=20N (C) N=200N 20kgd 20kgd N= 200N T=60N F= 120N 200N F2 = 60N 200N F1 = 60 N T=60N (D) All of the above [B] `Q.17 A force of 100 N is applied on a block of mass 3 kg as shown in figure. The coefficient of friction between the surface and block is 1/4. The friction force acting on the block is – 30o F = 100N (A) 15 N downwards (B) 25 N upwards (C) 20 N downwards (D) 20 N upwards [C] Q.18 A body A rests on B and friction coefficient between A and B is . Block M is placed on a frictionless surface. Then – A B m M (i) A B m M (ii) F F (A) the maximum possible value of F so that both bodies move together in case (i) is ( m + M)g (B) the maximum possible value of F so that both bodies move together in case is  mg (C) the maximum possible value of F so that both bodies move together in case (ii) is  (M + m) g (D) the maximum possible value of F so that both bodies move together in case (ii) is  Mg [C] Q.19 A bicycle is in motion. When it is not pedaled, the force of friction exerted by the ground on two wheels is such that it acts – (A) in the backward direction on the front wheel and in the forward direction on the rear wheel (B) in the forward direction on the front wheel and in the backward direction on rear wheel (C) in the backward direction on both the front and the rear wheels (D) in the forward direction on both the front and the rear wheels [C] Q.20 If a body of mass m is moving on a rough horizontal surface of coefficient of kinetic friction μ, the net electromagnetic force exerted by surface on the body is – (A) 2 mg 1 μ (B) μmg (C) mg (D) 2 mg 1 – μ [A] `Q.21 In the given figure, the coefficient of friction between m1 and m2 is μ and m2 and horizontal surface is zero – μ1 =  μ2 = 0 m1 F m2 (A) If F > μm1g, then relative acceleration is found between m1 and m2 (B) If F < μm1g, then no relative acceleration is found between m1 and m2 (C) It F > μm1g, then both bodies move together (D) (A) and (B) are correct [B]