NLM & Friction PDF

ALLEN® NLM & Friction 1...

ALLEN® NLM & Friction 1 NLM & FRICTION 7. A block of mass m slides along a floor while a 1. The coefficient of static friction between a force of magnitude F is applied to it at an angle wooden block of mass 0.5 kg and a vertical q as shown in figure. The coefficient of kinetic rough wall is 0.2. The magnitude of horizontal friction is µK. Then, the block's acceleration 'a' force that should be applied on the block to keep it adhere to the wall will be ______N. is given by : (g is acceleration due to gravity) [g = 10 ms–2] F 2. An inclined plane is bent in such a way that the x2 q vertical cross-section is given by y = where 4 y is in vertical and x in horizontal direction. If F æ F ö the upper surface of this curved plane is rough (1) - cos q - µ K ç g - sin q ÷ m è m ø with coefficient of friction µ = 0.5, the maximum height in cm at which a stationary F æ F ö (2) cos q - µ K ç g - sin q ÷ block will not slip downward is ________ cm. m è m ø 3. A person standing on a spring balance inside a F æ F ö stationary lift measures 60 kg. The weight of (3) cos q - µ K ç g + sin q ÷ that person if the lift descends with uniform m è m ø downward acceleration of 1.8 m/s2 will be_ N. F æ F ö [g = 10 m/s2] (4) cos q + µ K ç g - sin q ÷ m è m ø 4. A boy pushes a box of mass 2 kg with a force r ( ) F = 20iˆ + 10 ˆj N on a frictionless surface. If 8. Consider a frame that is made up of two thin massless rods AB and AC as shown in the the box was initially at rest, then ______ m is r displacement along the x-axis after 10 s. figure. A vertical force P of magnitude 100 N 5. As shown in the figure, a block of mass 3 kg is applied at point A of the frame. is kept on a horizontal rough surface of A 70° 1 coefficient of friction. The critical force 3 3 B to be applied on the vertical surface as shown at P an angle 60° with horizontal such that it does not move, will be 3x. The value of x will be 3 145° 1 [g = 10 m/s2; sin 60° = ; cos 60° = ] C 2 2 node06\B0BA-BB\Kota\JEE MAIN\Jee Main-2021_Subject Topic PDF With Solution\Physics\English\ NLM & Friction r Suppose the force is P resolved parallel to the 1 – µ= – m = Ö3 kg arms AB and AC of the frame. The magnitude 3Ö3 60° of the resolved component along the arm AC is 6. Two masses A and B, each of mass M are fixed xN. The value of x, to the nearest integer, is together by a massless spring. A force acts on ________. the mass B as shown in figure. If the mass A [Given : sin(35°) = 0.573, cos(35°) = 0.819 starts moving away from mass B with sin(110°) = 0.939, cos(110°) = –0.342] acceleration 'a', then the acceleration of mass B 9. A body of mass 2kg moves under a force of wil be :- (2iˆ + 3jˆ + 5k)N ˆ. It starts from rest and was at F B A the origin initially. After 4s, its new coordinates Ma - F MF (1) (2) are (8, b, 20). The value of b is __________. M F + Ma F + Ma F - Ma (Round off to the Nearest Integer) (3) (4) M M E 2 NLM & Friction ALLEN® 10. Two blocks (m = 0.5 kg and M = 4.5 kg) are 14. A steel block of 10 kg rests on a horizontal floor arranged on a horizontal frictionless table as as shown. When three iron cylinders are placed shown in figure. The coefficient of static on it as shown, the block and cylinders go down with an acceleration 0.2 m/s2. The normal 3 friction between the two blocks is. Then the reaction R' by the floor if mass of the iron 7 cylinders are equal and of 20 kg each, is ______ maximum horizontal force that can be applied N. [Take g = 10 m/s2 and µs = 0.2] on the larger block so that the blocks move together is ______ N. (Round off to the Nearest Integer) [Take g as 9.8 ms–2] m M F 11. A body of mass 1 kg rests on a horizontal floor with which it has a coefficient of static friction (1) 716 (2) 686 (3) 714 (4) 684 1 15. A particle of mass M originally at rest is. It is desired to make the body move by subjected to a force whose direction is constant 3 but magnitude varies with time according to the applying the minimum possible force F N. The relation value of F will be __________. (Round off to é æ t - T ö2 ù the Nearest Integer) [Take g = 10 ms–2] F = F0 ê1 - ç ÷ ú 12. A boy of mass 4 kg is standing on a piece of êë è T ø úû wood having mass 5kg. If the coefficient of Where F0 and T are constants. The force acts friction between the wood and the floor is 0.5, only for the time interval 2T. The velocity v of the maximum force that the boy can exert on the particle after time 2T is : the rope so that the piece of wood does not move from its place is ________N.(Round off (1) 2F0T / M (2) F0T / 2M to the Nearest Integer) [Take g = 10 ms–2] (3) 4F0T / 3M (4) F0T / 3M 16. A body of mass 'm' is launched up on a rough T T inclined plane making an angle of 30° with the F horizontal. The coefficient of friction between the R x T body and plane is if the time of ascent is half node06\B0BA-BB\Kota\JEE MAIN\Jee Main-2021_Subject Topic PDF With Solution\Physics\English\ NLM & Friction 5 T of the time of descent. The value of x is _____. 17. The coefficient of static friction between two blocks is 0.5 and the table is smooth. The 13. A bullet of mass 0.1 kg is fired on a wooden maximum horizontal force that can be applied block to pierce through it, but it stops after to move the blocks together is.......N.(Take: g = moving a distance of 50 cm into it. If the 10 ms–2) velocity of bullet before hitting the wood is 10 m/s and it slows down with uniform deceleration, then the magnitude of effective retarding force on the bullet is 'x' N. The value of 'x' to the nearest integer is ______. E  ALLEN® NLM & Friction 3 18. The boxes of masses 2 kg and 8 kg are 21. An object of mass 'm' is being moved with a connected by a massless string passing over constant velocity under the action of an applied smooth pulleys. Calculate the time taken by box force of 2N along a frictionless surface with of mass 8 kg to strike the ground starting from following surface profile. rest. (use g = 10 m/s2) The correct applied force vs distance graph will be: (1) (1) 0.34 s (2) 0.2 s (3) 0.25 s (4) 0.4 s 19. A car is moving on a plane inclined at 30º to the horizontal with an acceleration of 10 ms–2 parallel to the plane upward. A bob is suspended by a string from the roof of the (2) car.The angle in degrees which the string makes with the vertical is______. (Take g = 10 ms–2) 20. A block of mass m slides on the wooden wedge, which in turn slides backward on the horizontal surface. The acceleration of the block with respect to the wedge is : Given m = 8 kg, M = 16 kg. Assume all the surfaces shown in the figure to be frictionless. (3) node06\B0BA-BB\Kota\JEE MAIN\Jee Main-2021_Subject Topic PDF With Solution\Physics\English\ NLM & Friction (4) 4 6 3 2 (1) g (2) g (3) g (4) g 3 5 5 3 22. When a body slides down from rest along a smooth inclined plane making an angle of 30° with the horizontal, it takes time T. When the same body slides down from the rest along a rough inclined plane making the same angle and through the same distance, it takes time aT, where a is a constant greater than 1. The co- efficient of friction between the body and the 1 æ a2 – 1 ö rough plane is ç 2 ÷ where x =........... x è a ø E 4 NLM & Friction ALLEN® SOLUTION N'+Ma 1. Official Ans. by NTA (25) Sol. F.B.D. of the block is shown in the diagram Fr F Mg m N' = M (g – a) N Þ 60 (10 – 1.8) mg N' Þ 492 N Since block is at rest therefore fr – mg = 0 …(1) 4. Official Ans. by NTA (500) F–N=0 …(2) r Sol. F = 20iˆ + 10 ˆj fr £ µN r In limiting case r F 20iˆ + 10 ˆj fr = µN = µF …(3) a= = Þ 10iˆ + 5 ˆj m 2 Using eq. (1) and (3) r 1r \ µF = mg 0.5 ´ 10 1 ( ) \ s = at 2 = 10iˆ + 5ˆj ´ (10 ) 2 2 2 ÞF= = 25 N 0.2 Þ 50 (10iˆ + 5jˆ ) m Ans. 25.00 2. Official Ans. by NTA (25) \ Displacement along x-axis y Þ 50 × 10 Þ 500 m \ Ans. 500 h Sol. q x 5. Official Ans. by NTA (3) Fsin60° At maximum ht. block will experience Sol. – Fcos60° maximum friction force. Therefore if at this Ö3 g height slope of the tangent is tan q, then µN q = Angle of repose. N dy 2x x \ tan q = = = = 0.5 F 1 dx 4 2 F cos 60° = µN or = N... (1) 2 3 3 x2 Þ x = 1 and therefore y = = 0.25 m & N = sin 60° + 3g node06\B0BA-BB\Kota\JEE MAIN\Jee Main-2021_Subject Topic PDF With Solution\Physics\English\ NLM & Friction 4... (2) = 25 cm From equation (1) & (2) \ Answer is 25 cm (Assuming that x & y in the equation are given F 1 æF 3 ö = çç + 3g ÷÷ in meter) 2 3 3è 2 ø 3. Official Ans. by NTA (492) Þ F = g = 10 Newton = 3x N 10 So x = = 3.33 Sol. 3 Mg When lift is at rest N = mg Þ 60 × 10 = 600 N When lift moves with downward acceleration. In frame of lift pseudo force will be in upward direction. E  ALLEN® NLM & Friction 5 6. Official Ans. by NTA (4) 10. Official Ans. by NTA (21) m 1a1 + m 2a 2 3 Sol. acm = Sol. amax = mg = × 9.8 m1 + m 2 7 F = (M + m) amax = 5 amax F Ma + Ma B = 21 Newton = 2M 2M 11. Official Ans. by NTA (5) F F - Ma q aB = M Sol. N 7. Official Ans. by NTA (2) F sin q F cos q = mN N F F sin q + N = mg m mg Sol. q ÞF= F cos q cos q + m sin q 1 fk ´ 10 m mg mg Fmin = = 3 =5 N = mg – f sin q 1 + m2 2 F cos q – µkN = ma 3 F cos q – µk (mg – F sin q) = ma 12. Official Ans. by NTA (30) N F æ F ö a = cos q - µ K ç g - sin q ÷ T m m mN è ø Sol. T 8. Official Ans. by NTA (82) 9g A 70° N + T = 90 T = m N = 0.5 (90–T) 35° B 1.5 T = 45 Sol. P T = 30 55° 13. Official Ans. by NTA (10) Sol. v2 = u2 + 2as 145° æ1ö C 0 = (10)2 + 2 (–a) ç ÷ è2ø Component along AC a = 100 m/s2 = 100 cos 35°N F = ma = (0.1) (100) = 10 N node06\B0BA-BB\Kota\JEE MAIN\Jee Main-2021_Subject Topic PDF With Solution\Physics\English\ NLM & Friction = 100 × 0.819 N 14. Official Ans. by NTA (2) = 81.9 N » 82 N 9. Official Ans. by NTA (12) Sol. Ans. (12) Sol. r r F 2iˆ + 3ˆj + 5kˆ a= = m 2 = ˆi + 1.5 ˆj + 2.5kˆ r r 1r 2 Writing force equation in vertical direction t = ut + a t 2 Mg – N = Ma 1 ˆ Þ 70g – N = 70 × 0.2 = 0 + (i + 1.5jˆ + 2.5k) ˆ (16) 2 Þ N = 70 [g – 0.2] = 70 × 9.8 = 8iˆ + 12ˆj + 20kˆ \ N = 686 Newton b = 12 Note : Since there is no compressive normal from the sides, hence friction will not act. Hence option 2. E 6 NLM & Friction ALLEN® 15. Official Ans. by NTA (3) 18. Official Ans. by NTA (4) Sol. t = 0, u = 0 F F dv a = o – o 2 (t – T)2 = M MT dt Sol. v 2T æ Fo Fo 2ö ò dv = ò 0 t=0 ç M – MT 2 (t – T) ÷ dt è ø 2T 2T (m1g – 2T) = m1a – (1) é Fo ù Fo é t 3 2 2 ù T – m2g = m2(2a) V = ê tú – 2 ê – t T + T tú ë M ûo MT ë 3 û0 2T – 2m2g = 4m2 a – (2) 4F T m1g – 2m2g = (m1 + 4m2) a V= o 3M (8 - 4)g 4 g a= = g= 16. Official Ans. by NTA (3) (8 + 8) 16 4 1 10 Sol. ta = td a= m/s2 2 4 2s 1 2s 1 = ….(i) S = at 2 aa 2 a d 2 a a = gsin q + mg cos q 0.2 ´ 2 ´ 4 2 =t 10 g 3 = + mg t = 0.4 sec 2 2 19. Official Ans. by NTA (30) a d = g sin q - mg cos q g 3 = - mg 2 2 using the above values of aa and ad and putting 3 in eqution (i) we will gate m = Sol. 5 17. Official Ans. by NTA (15) node06\B0BA-BB\Kota\JEE MAIN\Jee Main-2021_Subject Topic PDF With Solution\Physics\English\ NLM & Friction mgsin 30° + ma Sol. tan ( 30 + q ) = mg cos30° F = 3a (For system) ….(i) 5 + 10 1 + 2 tan ( 30 + q) = = 5 3 3 1 fs max = 1a (for 1kg block) ….(ii) tan q + 3 = 3 µ×1×g=aÞ5=a F = 15N 1 1- tan q 3 3 tan q + 1 = 3 - 3 tan q 2 3 tan q = 2 1 tan q = 3 q = 30° E  ALLEN® NLM & Friction 7 20. Official Ans. by NTA (4) Sol. Let acceleration of wedge is a1 and acceleration of block w.r.t. wedge is a2 Þ F = 2N = (–ve) constant Þ Best possible answer is option (2) 22. Official Ans. by NTA (3) Sol. Ncos60° = Ma1 = 16a1 Þ N = 32a1 F.B.D. of block w.r.t wedge On smooth incline a = g sin30° 1 by S = ut + at 2 2 1g 2 g 2 S= T = T …….(i) ^ to incline 22 4 N = 8g cos 30° – 8a1 sin 30° Þ 32a1 = 4 3g - 4a1 3 Þ a1 = g 9 Along incline 8gsin30° + 8a1cos30° = ma2 = 8a2 node06\B0BA-BB\Kota\JEE MAIN\Jee Main-2021_Subject Topic PDF With Solution\Physics\English\ NLM & Friction 1 3 3 2g a2 = g ´ + g. = On rough incline 2 9 2 3 Option (4) a = g sin30° – µg cos 30° 21. Official Ans. by NTA (2) 1 by S = ut + at 2 Sol. During upward motion 2 1 ( ) S = g 1 - 3m ( aT ) …(ii) 4 2 By (i) and (ii) 1 2 1 4 ( ) gT = g 1 - 3m a2 T2 4 1 æ a2 - 1 ö 1 Þ 1 - 3g = 2 Þ g = ç ÷. a è a ø 3 2 Þ x = 3.00 F = 2N = (+ve) constant During downward motion E