Tiêu đề Friction (Pass.Q.) E.pdf ✅

PHYSICS Passage # 1 (Ques. 1 to 3) An external force F is applied at an angle  with the horizontal as shown on the block of mass 'm'. The coefficient of friction between block and wall is μ.  m F Rough (μ) Q.1 The minimum value of force F required to keep the block at rest is- (A) μcos mg (B) sin  μcos mg (C) sin  μcos mg (D) μtan  mg [C] Q.2 The maximum value of force F upto which block remains at rest is- (A) μcos mg (B) sin  μcos mg (C) sin  μcos mg (D) μtan  mg [A] Q.3 The value of force F for which friction force between block and wall is zero. (A) mg (B) sin  mg (C) cos mg (D) tan  mg [A] Passage # 2 (Ques. 4 to 6) ABCD is a horizontal surface on which a block of mass m = 20 kg is placed and coefficient of friction between this body and surface is μ = 0.5 as shown. y-axis is vertical and x-z plane is horizontal. m μ 0 x B C D z A y y The surface starts moving with an acceleration which depends on time as – a  = (3t i ˆ + 10 j ˆ + 4t k ˆ ) m/s2 (assume at t = 0 block is at rest with respect to the surface.) Q.4 The time 't' when block starts slipping on surface is – (A) 1 sec (B) 2 sec (C) 3 sec (D) None of these [B] Q.5 The value of total force applied by surface on block at t = 1 sec is – (A) 100 17 N (B) 200 2 N (C) 200 5 N (D) None of these [A] Q.6 The acceleration of surface when block starts slipping on it is – (A) 5 2 m/s2 (B) 10 m/s2 (C) 10 2 m/s2 (D) None of these [C] Passage # 3 (Ques. 7 to 9) A block on mass m1 = 1kg is placed on a wedge of mass m2 = 3kg as shown. The angle of inclination of the inclined plane of wedge is  = 30o. The wedge can move without friction on the horizontal surface. When the system is released from rest wedge is starts moving parallel to its incline surface as the string connecting it is ideal and horizontal and parallel to incline surface of wedge. The part of string connecting m is vertical and pulley is ideal. Assuming initially m1 is at rest with respect to m2 and tan < μ (g= 10m/s2 ). If μ = 0.9 then -
/////////////////////////////////////// \\\\\\\\\\\\\ m  m2  m o Q.7 The minimum value of m for which m1 just starts slipping on m2 is – (A) 4 kg (B) 6 kg (C) 8 kg (D) None of these [B] Q.8 For this minimum value of m acceleration of system is – (A) 6 m/s2 (B) 3 m/s2 (C) 4 m/s2 (D) None of these [A] Q.9 For this minimum value of m tension in string connecting m is (A) 30 N (B) 20 N (C) 24 N (D) None of these. [C] Passage # 4 (Ques. 10 to 12) an external force F is applied at an angle  with the horizontal as shown on the block of mass 'm'. The coefficient of friction between block and wall is . Q.10 The minimum value of force f required to keep the block at rest is – F  m rough (μ) (A) μcos mg (B) sin  μcos mg (C) sin  μcos mg (D) μtan  mg Sol. [B] F sin  + f2 = mg  F sin  + F cos  = mg  F = sin   cos mg Q.11 The maximum value of force F up to which block remains at rest is - (A) μcos mg (B) sin  μcos mg (C) sin  μcos mg (D) μtan  mg Sol. [C] F sin  = mg + f2  F sin  – F cos  = mg F = sin  sin  mg Q.12 The value of force F for which friction force between block and wall is zero - (A) mg (B) sin  mg (C) cos mg (D) tan  mg Sol. [B] F sin  = mg Passage - 5 (Q. 12 to 14) In an emergency situation while driving one may instinctively jam on the brakes, trying to come to a quick stop that is, to stop in the shortest distance. But with the wheels locked, the car skids or slides, to a stop, of tern out of control. To prevent this many newer automobiles have computerized antilock braking system (ABS). When the brakes are applied firmly and the car begins to slide, sensors in the wheels note this. In turn ABS momentarily releases the brakes and then varies the brake fluid pressure with a pumping action upto thirteen times a second so that the wheels will continue to roll without slipping. In the absence of sliding, both roiling friction and static friction act. In many cases, however, the force of rolling friction is small and only static friction need be taken into account. The
external force of static friction does no work to dissipate energy in slowing a car, it does determine whether the wheels roll or slide. Using Newton’s laws and a = μg, where is a stopping acceleration, then stopping distance x = 2μg V 2 Where V is initial speed at the moment of application of brakes. In case of only rolling friction coefficient μ is static and in case of sliding, it is kinetic in nature. Q.12 In a particular situation, a car without ABS needs 17.5 m as minimum stopping distance in sliding conditions. Under identical conditions, what minimum distance same car with ABS would required if μS = 0.75 and μk = 0.45 for these surfaces. (A) 17.5 m (B) 14 m (C) 10.5 m (D) 7 m [C] Q.13 Two identical cars, one with ABS and another without it, are stopped from same initial speeds by jamming the brakes. (A) Braking system of car with ABS would be more heated up. (B) Braking system of car without ABS would be more heated up. (C) Braking system of both cars would be heated up by same amount. (D) Insufficient information to come to conclusion. [A] Q.14 The ABS works to keep static friction f between road and wheel– (A) equal to limiting static friction (B) less than limiting static friction (C) less than kinetic friction (D) grater than limiting static friction [A] Passage – 6 (Q. 15o 1) All pulleys and strings shown in figure are light and there is no friction in any string and pulley. 0.5 8kg 6kg 0.4 m Q.15 Maximum mass 'm' to be hanged so that system remain in equilibrium - (A) 8 kg (B) 9.6 kg (C) 6.4 kg (D) 10 kg [A] Q.16 Mass 'm' to be hanged so that none of the block remains in equilibrium must satisfy - (A) m > 8 kg (B) m > 9.6 kg (C) m > 10 kg (D) m > 12 kg [B] Q.17 Acceleration of hanging block if mass of hanging block is 9 kg - (A) 31 9 m/s2 (B) 31 10 m/s2 (C) 41 10 m/s2 (D) 32 11 m/s2 [C] Passage – 7 (Q. 18 to 20) “Ramdin observe a 1350 kg sports car rolling along flat pavement in a straight line. The only horizontal forces acting on it are a constant rolling friction and air resistance (proportional to the square of its speed). Ramdin takes the following data during a time interval of 25 s: When its speed is 32 m/s, it slows down at a rate of –0.42 m/s2 , and when its speed is decreased to 24 m/s, it slows down at –0.30 m/s2 .” Q.18 Find the coefficient of rolling friction is– (A) 0.005 (B) 0.01 (C) 0.015 (D) None of these [C]
Q.19 The air drag constant is– (A) 0.18 N.s2 /m2 (B) 0.36 N. s 2 /m2 (C) 0.48 N. s 2 /m2 (D) None of these [B] Q.20 Retardation of car when its velocity is 30 m/s will be- (A) 0.387 m/s2 (B) 0.337 m/s2 (C) 0.37 m/s2 (D) 0.416 m/s2 [A] Passage # 8 (Ques. 21 to 23) Two blocks of masses 10 kg and 20 kg are moving together with a constant speed of 20 m/s and 10 kg block is not slipping on 20 kg when blocks are moving in region AB which is horizontal and smooth. A 10 kg  20 kg v B C The coefficient of friction between the blocks is 1 = 0.5. The horizontal surface BC is rough and coefficient of friction between 20 kg block and surface BC is 2 : (g = 10 m/s2 ) Q.21 When the blocks are moving on the smooth horizontal region AB of surface, then – (A) frictional force on 10 kg block is zero (B) frictional force on 10 kg block is 50 N (C) frictional force on 10 kg block is less then 50 N but non zero (D) None of the above [A] Sol. As there is no force on upper block of mass 10 kg and no tendency of slipping over 20 kg block therefore friction on it will be zero. Q.22 When this system of blocks enters the rough horizontal surface BC then still 10 kg block is not slipping on 20 kg block, then – (A) maximum value of 2 is 0.5 (B) maximum value of 2 is 0.75 (C) total kinetic friction is 100 N acting on 20 kg block for the above maximum value of 2 (D) total frictional force on 20 kg block is 150 N [A] Sol. Maximum acceleration or retardation upto which both blocks will move together is 1g = 5 m/s2  2(m1 + m2)g = (m1 + m2)a 2 = 0.5 Also kinetic friction on 20 kg is 2(m1 +m2)g = 150 N Q.23 If the value of 2 = 0.2, then in 15 seconds after the blocks have entered the rough surface - (A) The distance travelled by them in rough surface is 75 m (B) the distance travelled by them in rough surface is 100 m (C) Total force applied by 10 kg on 20 kg block is 102 N after 15 seconds (D) None of the above [B] Sol. As retardation is 5 m/s 2  v 2 = u 2 – 2aS 0 2 = 400 – 2 × 5 × S  S = 100 m Passage 9 : (Q. 24 to 26) An engineer is designing a conveyor system for loading lay bales into a wagon. Each bale is 0.25 m wide, 0.50 m high, and 0.80 m long (the dimension perpendicular to the plane of the figure), with mass 30.0 kg.  0.5 m 0.25 m Fig. The centre of gravity of each bale is at its geometrical center. The coefficient of static friction between a bale and the conveyor belt is 0.60, and the belt moves with constant speed. The angle  of the conveyor is slowly increased. At some critical angle a bale will tip (if it