SOLUIONS O CONCEPS CHAPE 6 1. Let ss of the block ro the freebody digr, 0...(1) velocity Agin 0 (fro (1)) g 4 g 4/g 4/10 0.4 he co-efficient of kinetic friction between the block nd the plne is 0.4. Due to friction the body will decelerte Let the decelertion be 0...(1) velocity 0 (fro (1)) g 0.1 10 1/s. Initil velocity u 10 /s inl velocity v 0 /s 1/s (decelertion) S v u 0 10 ( 1) 100 50 It will trvel 50 before coing to rest. 3. Body is kept on the horizontl tble. If no force is pplied, no frictionl force will be there f frictionl force Applied force ro grp it cn be seen tht when pplied force is zero, frictionl force is zero. 4. ro the free body digr, cos 0 cos..(1) or the block U 0, s 8, t sec. s ut + ½ t 8 0 + ½ 4/s Agin, + sin 0 cos + sin 0 [fro (1)] (g cos + g sin ) 0 10 cos g sin p o 10 ( 3 / 3) 10 (1/) 4 ( 5 / 3 ) 1 1/ ( 5 / 3 ) 0.11 Co-efficient of kinetic friction between the two is 0.11. 5. ro the free body digr 4 4 + 4g sin 0 (1) 4g cos 0...() 4g cos Putting the vlues of is & in equn. (1) 4 4 0.11 4g cos + 4g sin 0 4kg 4N 4 4 0.11 4 10 ( 3 / ) + 4 10 (1/) 0 4 4 3.81 + 0 0 5 /s or the block u 0, t sec, 5/s Distnce s ut + ½ t s 0 + (1/) 5 10 he block will ove 10. 6.1
6. o ke the block ove up the incline, the force should be equl nd opposite to the net force cting down the incline + g sin 0. (9.8) 3 + I 9.8 (1/) [fro (1)] 3.39 + 9.8 13N With this iniu force the body ove up the incline with constnt velocity s net force on it is zero. b) Net force cting down the incline is given by, g sin 9.8 (1/) 3.39 6.41N Due to 6.41N the body will ove down the incline with ccelertion. No externl force is required. orce required is zero. 7. ro the free body digr g 10/s, kg,, 0. cos - sin 0 cos + sin...(1) And sin + cos 0 sin + ( cos + sin ) cos 0 sin + cos + sin cos 0 ( sin cos) ( sin cos) 10 (1/ ) 0. 10 ( 8. ss of child cos 45 0 0. (1/ ) ( 3 / ) 3 / ) cos 45 /v...(1) 13.464 17.7N 17.5N 0.76 Net force cting on the boy due to which it slides down is sin 45 - sin 45 - cos 45 10 (1/ ) 0.6 10 (1/ ) [(5/ ) 0.6 (5 / )] ( ) orce ccelertion ss ( ) /s 9. Suppose, the body is ccelerting down with ccelertion. ro the free body digr cos 0 cos...(1) + sin 0 (sin cos ) g (sin cos ) or the first hlf t. u 0, s 0.5, t 0.5 sec. So, v u + t 0 + (0.5)4 /s S ut + ½ t 0.5 0 + ½ (0/5) 4/s...() or the next hlf etre u` /s, 4/s, s 0.5. 0.5 t + (1/) 4 t t + t 0.5 0 6. Chpter 6 (body oving down) (body oving us) 45
4 t + 4 t 1 0 4 16 16 1.656 0.07sec 4 8 ie tken to cover next hlf eter is 0.1sec. 10. f pplied force i contct force frictionl force norl rection tn / Chpter 6 When, is the liiting friction (x friction). When pplied force increse, force of friction increse upto liiting friction () Before reching liiting friction < tn 11. ro the free body digr tn tn 1 + 0.5 0.5 g 0...(1) + 1 + 1 0...() + 1 1 0 + 1 1...(3) ro () & (3) + 1 1 1 A 1kg 0. B 1kg 0. i 0.5kg f Liiting riction 0.5g 0.5g 1 Eqution () becoes + + 1 1 0 + 1 0 1 + 0.g +...(4) Eqution (1) becoes 1 + 0/5 0.5g 0 1 0.5g 0.5 0.5g 0.5...(5) ro (4) & (5) 0.g + 0.5g 0.5 1 1 A 1g 1 1g 0.05 10 0.04 I 10 0.4/s 1.5 ) Accln of 1kg blocks ech is 0.4/s b) ension 1 0.g + + 0.4.4N c) 0.5g 0.5 0.5 10 0.5 0.4 4.8N 1. ro the free body digr 1 + 1 16 0 1 (g) + ( 15) 0 1 15/0 0.75 1 + 4 0.5 + 16 4g sin 0 (0 3 ) + + 16 0 0 kg 1 0.5 /s 4kg 0 3 1 17.3 0.057 0.06 Co-efficient of friction 1 0.75 & 0.06 0.5 1 16N 4g 4 0.5 16N 6.3
Chpter 6 13. A 15kg ro the free body digr + 15 15g 0 ( 1 + 5+ ) 0 1 5g 5 0 15g 15...(i) (5g + 5 + 5 + ) 0 1 5g + 5 (iii) 5g + 10 + (ii) ro (i) & (ii) 15g 15 5g + 10 + 0. (5g) 5 90 3.6/s Eqution (ii) 5 10 + 10 3.6 + 0. 5 10 96N in the left string Eqution (iii) 1 5g + 5 5 10 + 5 3.6 68N in the right string. 14. s 5, 4/3, g 10/s u 36k/h 10/s, v 0, v u 0 10 10/s s 5 ro the freebody digrs, cos 0 ; g 10/s cos.(i) ; 4/3. Agin, + sin - 0 + sin cos 0 + g sin cos 0 10 + 10 sin - (4/3) 10 cos 0 30 + 30 sin 40 cos 0 3 + 3 sin 4 cos 0 4 cos - 3 sin 3 4 B C 15kg 1 sin 3 + 3 sin 16 (1 sin ) 9 + 9 sin + 18 sin A 15 15g B r5g 5g 1 5 1 5g the x. ngle velocity sin 18 sin 1 (0.8) 16 18 4(5)( 7) 5 18 3 50 14 0.8 [king +ve sign only] 50 Mxiu incline is 16 15. to rech in iniu tie, he hs to ove with xiu possible ccelertion. Let, the xiu ccelertion is 0 g 0.9 10 9/s ) Initil velocity u 0, t? 9/s, s 50 s ut + ½ t 50 0 + (1/) 9 t t 100 10 sec. 9 3 b) After overing 50, velocity of the thelete is V u + t 0 + 9 (10/3) 30/s He hs to stop in iniu tie. So decelertion i 9/s (x) 6.4
Chpter 6 (x frictionl force) g 9 / s (Decelertion) u 1 30/s, v 1 0 1 1 v u 0 30 30 10 t sec. 3 16. Hrdest brke ens xiu force of friction is developed between cr s type & rod. Mx frictionl force ro the free body digr cos 0 cos...(i) nd + sin ) 0 cos + sin 0 g cos + 10 (1/) 0 (ii) 5 {1 ( 3 )} 10 ( 3 / ).5 /s When, hrdest brke is pplied the cr ove with ccelertion.5/s S 1.8, u 6/s S0, velocity t the end of incline V u s 6 (.5)(1.8) 36 64 10/s 36k/h Hence how hrd the driver pplies the brkes, tht cr reches the botto with lest velocity 36k/h. 17. Let,, xiu ccelertion produced in cr. [or ore ccelertion, the tyres will slip] g 1 10 10/s or crossing the bridge in iniu tie, it hs to trvel with xiu ccelertion u 0, s 500, 10/s s ut + ½ t 500 0 + (1/) 10 t t 10 sec. If ccelertion is less thn 10/s, tie will be ore thn 10sec. So one cn t drive through the bridge in less thn 10sec. 18. ro the free body digr 4g cos 4 10 3 / 0 3...(i) + 4 P 4g sin 0 0.3 (40) cos + 4 P 40 sin 0 0 (ii) P + + 1 1 g sin 0 (iii) 1 g cos 10 3 / 10 3...(iv) Equn. (ii) 6 3 + 4 P 0 0 Equn (iv) P + + 3 10 0 ro Equn (ii) & (iv) 6 3 + 6 30 + 3 0 6 30 8 3 30 13.85 16.15 P 16.15.69.7/s 4g g 6 b) cn be solved. In this cse, the 4 kg block will trvel with ore ccelertion becuse, coefficient of friction is less thn tht of kg. So, they will ove seprtely. Drwing the free body digr of kg ss only, it cn be found tht,.4/s. 6.5 P kg 4kg 1 1
Chpter 6 19. ro the free body digr M 1 M 1 1 M 1 M 1 g cos M g cos...(i)...(ii) + M 1 g sin 1 1 0 M M + 0...(iv)...(iii) Equn (iii) + M 1 g sin M 1 M 1 g cos 0 Equn (iv) M g sin + M + M g cos 0...(v) Equn (iv) & (v) g sin (M 1 + M ) (M 1 + M ) g cos (M 1 + M ) 0 (M 1 + M ) g sin (M 1 + M ) g cos (M 1 + M ) g(sin cos ) he blocks (syste hs ccelertion g(sin cos ) he force exerted by the rod on one of the blocks is tension. ension M 1 g sin + M 1 + M 1 g sin M 1 g sin + M 1 (g sin g cos ) + M 1 g cos 0 0. Let p be the force pplied to t n ngle ro the free body digr + P sin 0 P sin + p cos...(ii)...(i) Equn. (i) is ( P sin ) P cos 0 sin P cos M 1p sin cos M g P Applied force P should be iniu, when sin + cos is xiu. Agin, sin + cos is xiu when its derivtive is zero. d/d ( sin + cos ) 0 cos sin 0 tn 1 So, P sin cos sec (1 tn / cos sin cos cos cos 1 sec 1 tn sec 1 tn Miniu force is 1 t n ngle tn 1. 1. Let, the x force exerted by the n is. ro the free body digr + Mg 0 Mg 1 0 1 + And 1 0...(i)...(ii) 1 1 6.6
Chpter 6 ( + ) 0 [ro equn. (ii)] 0 (Mg + ) 0 [fro (i)] (1 + ) Mg + (M )g 1 Mxiu force exerted by n is (M )g 1. 1N kg 1 1 4kg 0. 1 1N 4 1 1 g 0 1 10 0 + 0. 1 1 0 4 1 1 0 4 1 1 0. (0) + 0.(0) 1 4 1 4 1 4 8 1 1/s 4/s kg block hs ccelertion 4/s & tht of 4 kg is 1/s 1N kg 4kg 1 g 1 4 1 4g g 1 1 3. g 4g g (ii) 1 g 0 4 + 0. 10 1 0 M 1 0 4 + 4 1 0. (0) 4 4 8 /s /s 1 0. A kg 1 0.3 B 3 kg 1 0.5 C 7 kg 10N 14N g 1 10N 15N 10N 3g 5g ) When the 10N force pplied on kg block, it experiences xiu frictionl force 1 kg (0.) 0 4N fro the 3kg block. So, the kg block experiences net force of 10 4 6N So, 1 6/ 3 /s But for the 3kg block, (fig-3) the frictionl force fro kg block (4N) becoes the driving force nd the xiu frictionl force between 3kg nd 7 kg block is (0.3) 5kg 15N So, the 3kg block cnnot ove reltive to the 7kg block. he 3kg block nd 7kg block both will hve se ccelertion ( 3 ) which will be due to the 4N force becuse there is no friction fro the floor. 3 4/10 0.4/s 6.7
Chpter 6 A kg B 3 kg C 7 kg 10N 4N 15N g 3kg 3g 10N 5g b) When the 10N force is pplied to the 3kg block, it cn experience xiu frictionl force of 15 + 4 19N fro the kg block & 7kg block. So, it cn not ove with respect to the. As the floor is frictionless, ll the three bodies will ove together 1 3 10/1 (5/6)/s c) Siilrly, it cn be proved tht when the 10N force is pplied to the 7kg block, ll the three blocks will ove together. Agin 1 3 (5/6)/s 4. Both upper block & lower block will hve ccelertion /s 1 1 M 1 1...(i) 1 0 1 0 0...(ii) + [putting ] 1 1 1 1 b) 0 (i) M 0 [ 1 ] M + Putting vlue of in (i) f M 0 () (M + ) [Putting ] 4 (M + ) M 1 5. Both blocks ove with this ccelertion in opposite direction. 1 1 1 + 0 1 (g )...(i) 1 0 ( ) Agin, 1 0...(ii) M 1 1 1 ( ) 6.8
Chpter 6 {( )} u( ) 0 + + 0 (g ) b) Accelertion of the block be 1 1 1 1 1 1 1 1 1...(i) 1 M 1 0 1 1 0 1 + M 1 t + 1 0...(ii) ( ) + M 1 + M 1 Subtrcting vlues of &, we get ((g )) ( + M 1 ) + 1 0 4 4 + 1 + M 1 1 Both blocks ove with this ccelertion but in opposite directions. 6. 1 + QE 0 1 QE...(i) 1 0 ( QE) 0 + QE 0 () - 1 0 6.9 (g ) M 1 ( QE) QE 1 Now eqution (ii) is + QE + QE 0 1 + QE 0 QE QE 1 ( QE) Mxiu horizontl force tht cn be pplied is ( QE). 7. Becuse the block slips on the tble, xiu frictionl force cts on it. ro the free body digr 0 But the tble is t rest. So, frictionl force t the legs of the tble is not 1. Let be f, so for the free body digr. o 0 o. otl frictionl force on tble by floor is. 8. Let the ccelertion of block M is towrds right. So, the block ust go down with n ccelertion. M 1 1 (BD-1) 1 As the block is in contct with the block M, it will lso hve ccelertion towrds right. So, it will experience two inerti forces s shown in the free body digr-1. ro free body digr -1 M 1 1 Mg (BD-) 1 1 M E QE
Chpter 6 1 0 1...(i) Agin, + + 1 1 0 ( 1 ) ro free body digr- + 1 1 + 0 + 1 + Mg (ii) [Putting the vlue of 1 fro (i)] ( 1 ) + 1 + Mg [Putting the vlue of fro (ii)] Mg + (iii) Agin, for the free body digr - + M 0 MA A (Mg + ) 0 (M + ) + (Mg + ) ro eqution (ii) nd (iv)...(iv) ( + 1 ) (M + ) + (Mg + ) (M + )g (M + + 4 + 1 ) [ (M )]g M [5 ( )] 1 9. Net force *(0 + (15) (0.5) 40 5 0 5N tn 0/15 4/3 tn 1 (4/3) 53 So, the block will ove t n ngle 53 with n 15N force 30. ) Mss of n 50kg. g 10 /s 6.10 [Putting the vlues of 1 nd fro (i) nd (iii)] rictionl force developed between hnds, legs & bck side with the wll the wt of n. So he reins in equilibriu. He gives equl force on both the wlls so gets equl rection fro both the wlls. If he pplies unequl forces should be different he cn t rest between the wlls. rictionl force blnce his wt. ro the free body digr 40 10 + 40g 40 10 50N 0.8 b) he norl force is 50 N. 31. Let 1 nd be the ccelertions of nd M respectively. Here, 1 > so tht oves on M Suppose, fter tie t seprte fro M. In this tie, covers vt + ½ 1 t nd S M vt + ½ t or to to seprte fro M. vt + ½ 1 t vt + ½ t +l...(1) Agin fro free body digr M 1 + / 0 M 1 (/) (/) 10 1 5 < Agin, M+ M + (M + )g (/) 0 M + (M + )g 0 (M+)g M Mg Mg M Putting vlues of 1 & in eqution (1) we cn find tht 4l (M ) g 1 M velocity l 40g 1 1