laws of motion & friction - wavesiitjee.com solution.pdf · laws of motion & friction...

LAWS OF MOTION & FRICTION (Solution) Target IIT-JEE/AIIMS 2019

3LAWS OF MOTION & FRICTION

Physics Secrets by Shiv R. Goel(meant only for WAVES classroom students)

SOLUTIONEXERCISE-1 (GRADE- I)

1. B2. AB

sin = 54

and cos = 53

3

45

AT

= 1000 cos i + 1000 sin )j( =

Nj800i600

AB TT

= j800i600

3. DAction rection pair act on different bodies and areof some type.

N

N

mg

mg

Earth

Normal force exerted by table on book is thereaction force of normal force exerted by the bookon the table.Normal force exerted by the ground on the tableis the reaction force of normal force exerted by

the table on the ground.4. C

Fdrag is opposite to the direction of motion'ma' is not a force, acceleration is the effect of netforce on a body.gravitational force acts downwards.As the helicopter moves with constant velocitythere force acceleration is zero forces must balanceeach otherThus Frotor acts such that is balnces the cesultantof Fdrag & Fgravitational

5. CString will break easily if the tension in the stringis more.2cos = W

w

T

T T

T =cos2

w

as increases, cos decreases, T increasesdrawing FBD for the lower half

6. D

60° TB

100

TA

TB cos 60° = 100 TB = 200 NTB sin 60° = TA

TA = 100 3 N7. B8. A

2T sin = 10


0.1T T

25m

10m

tan = 251.0

~ sin

2T × 251.0

= 10

T = 1250 N ]

9. D10. C11. B12. B13. D14. A15. A16. C

a = g sin

1001g = 0.1 m/s2

17. Tx = 2T

; TTy = 23T

Tx= 23T

; TTy = 2T

Wx = 0; Wy = –W; WX = 2W

; WWy =

23W

; Nx = 2N

;

Ny = 23N

; NX = 0 ; Ny = N]

18. B

(B).50N

100N

70N

25N

T

70N

Short and thick

a = 5100–120

95 – T = 2.5 × 4= 4 m/s2

T = 85 N19. A

20. P|| = 5P3

,W|| = 5W4

,

T|| = T, P = 5P4

,W = 5W3

, TT = 0

21. CCheck all options.For example, see option(C).Fx = 2(–1)2 – 2 = 0Fy = (–1) – (–1) = 0Fz = 5(–1)2 – 2(2)2 – 3(–1) = 0

22. DM P

m

P = (m + M)aM T T = Ma

23. C

a =21

21mmmm

g

5 = 2m2m

× 10

5m + 10 = 10m – 2030 = 5 mm = 6 kg

24. BF F = ma

F

F cos = ma N +

F sin = mg25. B26. B27. B28. C29. C30. C31. C32. C33. D34. B35. A36. C37. C,D38. C39. [(A) — Q, (B) — P, (C) — R, (D) — S]


When block will slip over the wedge, wedge willmove in right side. Thus acceleration of blockwill make more than angle from the horizontal.Normal force will be normal to the surface of thewedge & block. Since wedge move in right side,

40. A

f = 0.9

F

in limiting condition,f = × F cos

F sin = f = Fcos tan = = 0.9 = tan–1 (0.9)

41. B

37º

N

= 1/3100 g

100kgT

T

25 g

T

a

N + T sin37º = 100gN = 100g – T sin 37ºT – 25 g = 25a

T = 25 g + 25 aFor block to start moving

T cos 37º = × (100 g – T sin 37º)

(g + a ) =2s/m

3ga

53

31

5425

)g100(31

42. D43. B

(B)32

3

K'

for a particular meterial of spring

1K K= constant

32K

= K ×

K' = 2K3

44. B

3 kg 2 kg 10N2ms–2

T T 2 m/s2

for 2 kg mass10 – T = 2 × 2

T = 6 Nfor 3 kg mass

T = 3 × a a = 2 m/s2

45. D

.(D)

N N

mg sin30º mg cos30º = 0.5

30º

N = mg cos30º f = mg cos30º

contact force = 22 Nf =

23102

231025.0

2

= Nt55


46. A

j4iF

F

1 kg

= 0.3

Drawing FBD of block,

i + 4 jN

1 g = 10 Nt

f 0.3 N

In vertical direction,N + 4 = 10N = 6f 0.3 × 6

As the horizontal component of F is 1 Nt, (< 1.8Nt) friction is static which is also equal to 1 Ntin opposite direction.

47. B

. 37º

N

= 1/3100 g

100kgT

T

25 g

T

a

N + T sin37º = 100gN = 100g – T sin 37ºT – 25 g = 25a

T = 25 g + 25 aFor block to start moving

T cos 37º = × (100 g – T sin 37º)

48. D49. B50. A51. C52. A53. D

54. B55. D56. A57. B58. A59. A,B,C60. A61. B,C62. A,B,D63. B,D64. B,C,D65. C,D66. B67. [ (A) P,S (B) P,R (C) Q,R]

mg = 75 = 0.1F = 100 N

(A) = 37°

m

F

F = 100 Nm = 7.5 kgµ = 0.1

F cos

F sinmg

N

Fsin = 60 NN = Fcos =80 N fA 8 Nf = 8 N (upwards) , Kinetic friction acts .

(B) = 45°N = 250 fB 25Fsin = 250 ,

f = 75 – 250 = 25 ( 3 – 22 )f = 5N upwards(static friction acts)

C) = 53°Fsin = 80NN = 60 fC 6N,f = 5N downwards (static friction acts )


EXERCISE-1 (GRADE- II)1. A

o45 o45

mgm

1T1T

1TM

2T

Mg

2T1 cos 45o = mg 1mgT

2

T2 sin = Mg + T1cos45o

T2 sin = Mg +mg2

…..(1)

T2 cos = T1 sin 45o

T2 cos =mg2

….(2)

(1) (2)2Mtan 1m

2. (D) For A : T = mgFor B: T + mg cos = mg sin mg mg cos mgsin

+ cos = sin Squaring both sides2 + 2 cos2 + 2 2 cos = 1 – cos2 2 2 2 2( 1) cos 2 cos 1 0

2 2 2 2

2

2 4 4( 1)( 1)cos

2( 1)

=2

2

11

=

2

2

11

(ignoring –ive sign)

2

2

1cos1

3.B 3T = 50g + 25g T = 25g = 250 N

T TT

50 g25g

A,CBefore burning BC, the Free Body Diagramsare shownT

2 = T

1 + m

2g —(1)

kx = T2 = m

1g ——(2)

where x is extension in the spring.Just after buring, T

1 will become

zero , but T2 will remain same

T2 – m

2g = m

2a a =

1 2

2

(m m )gm

8.A,C 2T cos = F FT

2cos

FT

TIf increases, cos decreases and hence tensionincreases.

Now T > F F F

2cos

1 > 2 cos cos < 1/2 > /3

9.A,C Tension will be equal to apparent weight of 5 kgblock, which is 5(g + a) = 5 [10 + 2] = 60 N

10.B 2TvA = 3TvB A B3v v2

A 03v v2

A/B A Bv v v 00

3vv

2 = 0v

211.B T1 = 2T2

2T 2T

1 1T m g

B

2 3

12 3

2m m gm g 2

m m

1 2 3

4 1 1m m m

LAWS OF MOTION & FRICTION (Solution) Target IIT-JEE/AIIMS 201913. (c)

2 2F f N , f varies from 0 to N, where N = Mg

So F varies from N to 2 2( N) N

2 2N F ( N) N

2Mg F Mg 1

15.C Given 1 2 3 4ˆF F F F 100 3i

..(1)

2 3 4ˆF F F 100 1 i

..(2)

1 3 4ˆF F F 100 24 j

..(3)

(1)×2 – (2) – (3) we get

1 2ˆ ˆF F 700i 2400j

1 2F F ˆ ˆa 7i 24jm

2 2 2a 7 (24) 25 m / s 17.(B) Let f is the resistive force.

fT f

T

mg

A B

for A: T = f , for B: T + f = mgfrom above T = mg/2

19.B 70 + 5g – 100 = 5a

a

100 N

5 g

70 N

2.5 ga

T

70 N

a = 4 m/s2

2.5 g + 70 – T = 2.5 a T = 85 N

20.(B) Let tension in the spring is T, thenT

a

T3 kg 2 kg

22 m/ s

10 N

10 – T = 2 × 2 T = 6 NT = 3a a = T/3 = 6/3= 2 m/s2

21.(A) 21S at2

2 1t

a

22

1

at2t a

1 gsin g cos4 gsin

3 tan4

22.(a) (i) A (ii) A (b) (i) B (ii) AIn part (a), both blocks move togetherIn part (b), blocks move separately.

27.(I) A (II) D (III) BFriction on 2 kg and 4 kg will be limiting, but on 6 kg

friction will be less than limiting.T + 8 + 4 = 20 T = 8 N

4 kg 2 kg20 NT

2f 8 N 1

f 4 N

EXERCISE#2 (ONE OR MORE THAN ONE )

1.(BD) (fs)max = 0.25 × 10 = 2.5 NFmax = 2NSo block does not move

a = 03. (BCD) Minimum value of force F to keep block at rest

F = mg (F = N) 0.5 F = 0.1 × 10 F = 20 N

That means if F 20 N friction will be equal tomg (10 N)

FN

mg

N

F = NAs block can not move

perpendicular to the wall

5. (BCD) 30 = µsmg30 = 0.3 m (10)m = 10 kgSo total mass must be greater than 10 kgSo m 4 kg


EXERCISE#4 (SUBJECTIVE)

1 considering equilibrium of part PQ of the ropetension at point P should be equal to the weigth ofpart PQ.

x L

T

xgLm

LmxgT ]

Q.2 Applying newton's II law on whole system

F – 40 – 21

× 10 = (4 + 21

)a

54 – 40 – 5 = 4.5a9 = 4.5aa = 2 m/s2

Applying Newton's II law on lower half

T – 21

(45) = 25.4

× a

B

T

T = 245

+ 25.4

× 2 = 245

+ 4.5

= 22.5 + 4.5 = 27 N ]Q.3 Applying Newton's II law on the two blocks :

8 Fx

8kg

T

N

Fx – T = 8a

2

T

2kg

T – 2g = 20

Adding 1 & 2Fx – 2a = 10a

for a > 0,Fx > 20(i) & (ii)Fx – T – 4T + 8g = 0

T = 5g8Fx

for string toblock , T 0Fx – 8 g

[Ans. (a) Fx > 20 N (b) Fx – 80 N]

Q.4 [Ans. T = 7Mg8

]

Sol. As pulleys are massless

T1 = 2F ....(1)T2 = 2T1 = 4F ....(2)T = 2T2 = 8F ....(3)For equilibrium of mT2 + T1 + T = Mg4F + 2F + F = Mg7 F = Mg

F = 7Mg

putting in (3)

T = 7Mg8

]

Q.5

F

200N

2F

F

(a) Considering the whole system net upward force= 2Fthe maximum acceleration of man would be whenthe upper rope has maximum tension


Maximum upward force, 2F = 300 Ndownward force w = 200 NApplying Newton's II law300 – 200 = 20 amax

amax = 20200–300

= 5 m/s2

(b) (i) when painter is at rest2 F = wF = 100 N

(ii) When painter moves with an acceleration,a = 2m/s2

2F – w = 20 × 2

F = 2240

= 120 N

Q.6 [2 sec]

m1 m2

F = 30t N

4kg 1kg

3F

3F

3F

3F2

m1

40

N 3F2

t103F

t203F2

for m1 to lose contact with the floor N = 0

3F2

= 40 = 20 t

t = 2 sec ]

Q.7 [Ans : 6 N]

3kg2kgP = 10 N A B

due to constraint both blocks will move with sameacceleration

T

netsys M

Fa a = 510

= 2 m/s2

AN

a

for block AP – N = ma × a10 – N = 2×2N = 6 N

Q.8 [3 : 1]

[Sol.5kg 2kg 1kg

AB

CF = 16N

N2N1N

The three blocks will move with same acceleration,

asys = 816

= 2 m/s2

5kg16N N1

16 – N1 = 5 × aN1 = 6 N

1kgN2 N2 = 1 × a

3NN

2

1

= 2 N ]

Q.9 [Ans 5N, 16/31 kg ]

[Sol.37°

1kg3 m/s2

A

Ba

37ºa

AN2

53º

Applying Newton's II law to A in verticalMa g – N1 cos37º = MA = × a

.......(i)by constraintsaA cos 37º = 3 sin37º

.......(ii)

37ºN 1

3m/s2

M gB


applying Newton's II lawN1 cos53º = mB × 3N1 × 1 × 3 ..........(iii)from equation (i) & (ii) cos can find mA and a

Q.10 [Ans. (a) 31

mm

2

1 (b) a = 3/4 m/s2 ]

F = m1 × 3F = m2 × 1

(a) dividing the equation

31

mm

2

1

(b) If the masses are combinedF = (m1 + m2)a=m1 × 3substituing the value of m2

m1 a + 3m1 a = 3 m1

a =43 m/s2 ]

Q.11 [Ans : m'm]

[Sol. m m'

force on both the masses is same i.e. kxkx = makx = m'a'

'mm

'aa ]

Q.12 [Ans : 0.5 m]

[Sol.

= 30º

K = 40 N/m

m1

m2

m2

2 kg

3 kg

m1

T

20

30

T

in equilibrium conditionfor m1 ,m1 g sin30 + kx = T10 + 40x = T .........(i)for m2m2g = T Þ T = 30 ..........(ii)from (i) & (ii) x = 0.5 m

Q.13 [Ans: x1 : x2 : x3 : 15 : 18 : 10]

[Sol.

20 20

Tkx = T1

a1

for constant extension a of the blocks is same;kx1 – 20 = 2a120 – kx = 2a1

a1 = 0, kx1 = 20

x1 = k20

m

(b)

kxa2

2 kx2

3020

30 – kx2 = 3a2kx2 – 20 = 2a2

solving,a = 2 m/s2

x2 = k24

m

(c)

kx3

a3

kx3

10 20

kx3 – 10 = 1a320 – kx3 = 2a3

Solving

a3 = 310

m/s2

x3 = k340

]


Q.14 (a) 30° (b)23

]

[Sol.

g sin

(a) S t r i n g w o u l d m a k e a n g l e with thevertical if a = g sin.

a = g sin = 630

= 5 m/s2

sin = 21

= 30º

(b)

0.1a(pseudo force)

T

0.1g

for equilibriumT sin = 0.1a cosT = 0.1 × 15 × 3

=23

N ]

Q.15 [Ans. (M + m1 + m2)

g

mm

1

2 ]

[Sol. m2

m1

MF

for alll blocks to move togetherF = (M + m1 + m2) a ......(i)

Tm1

T = m1 a .......(ii)Sol. Note that the force exerted by the string accelerates

m1

m2

T

N

m g2

T = m2 g ........(iii)

Solving (ii), (iiii)

a =1

2m

gm

Substituting a in (i)

F = (m1 + m2 + M)1

2m

gm]

Q.16 [Ans 55]

[Sol. 37°

5 m/s2

N25N

50N

37º

Considering motion of the block w.r..t theinclined plane

Pseudo force on block FP = 25 NApplying Newton's law on the block in directionperpendicular to the inclined surface,

N = 25 sin 37º + 50 cos 37ºN = 15 + 40N = 55 Newton

By Newton's Third law,this force exerted by inclined plane on the blockis equal to the force exerted by the block on theinclined plane ]

Q.17 [Ans : 0.56]

[Sol.mg sin37º

37°mgcos37º

NN = mgcos37º

a1 = mº37cosmgº–37sinmg

if friction was absenta2 = g sin37º


The pi g starts from rest in both theeases &distance travelled is same,

211ta

21

= 222ta

21

also t = 2t2

º37cos4º37sin3

]

Q.18 [Ans : (a) 10.8 N ; (b) 0.33 m/s2 ; ]

[Sol.= 37º

a

N

3.5gk =0.25 15 N

3.5kgfk

Resolving the froces in vertical and horizontal andapplying Newton's II law(a) N = 3.5 g + 15 sin 37º = 35 × 9.8 + 9= 43.3fk = N= 0.25 × 43.3= 10.8 N15 cos 37º – fk = 3.5 a

(b) a = 5.38.10º–37cos15

= 0.33 m/s2 ]

Q.19[Ans : 1/3]

[Sol. a = g/3

37º

N

mg cos37ºmg sin37ºmg

N

Applying Newton's Law in direction perpendicularand parallel to the plane,

N = mg cos 37º

a = mº37cosmgº–37sinmg

3g

= g53

– g54

= 31

]

Q.20 [Ans : 2000 N]

[Sol.A

90 kgB 10kgF

= 0.5

acceleration of the blocks, a = 100F

FBD of B,

a

N

f

10 gApplying Newton's II Law in horizontal for blockB,

N = 10 × 100F

For limiting condition f = Nf = 10 g

N 10g

0.5 × 10F 10 g

F 2000 N. ]

Q.21 [Ans1 kg]

[Sol. m1kg37°

=0.5A T

Tf

1g cos37º1g sin37º


for block A to slide up with constant velocity,a = 0

mg = T.......(i)T + f = 1g sin 37º.......(ii)N = 1g cos 37º.......(iii)f N.......(iv)

Solving the equation we getm = 1kg ]

Q.22 [Ans. 332

m/s2]

Sol.

A

B

NT

T2T

2T

4T

By constraints aB = 4aA

A a

4T

50

B

4a

T

10for block BN = 10 g fk = ux 10 × g50 – 4T = 5a ... (1)T – 10 = 10 (4a) ... (2)

a = 332

m/s2

Q.23N N NNN N

f

f1 f1

f1f1

f

From FBD,

For center book2f1 = mg

f1 N N > 2mg

N > 2mg

For side bookf – f1 = mg

f = 2mg3

N

N 2mg3

Minimum N = 2mg3

(Greater of these value)

]

Q.24 [Ans 0, i10 ]

Sol.

5Kg5KgA

B

F = 50Nx

4Kg

=0.8g=10m/s2

a

a

T

B

40

Considering acceleration of blocks in the directionof applied force F,for block B,T – 40 = 4a.......(i)

N

50

50T

fA


for block A,Applying Newton's II Law

50 – T – f = 5a.........(ii)Adding (i) & (ii)

f = 10 – 9a..........(ii)i f a 0,friction f is kineticfk = N= 0.8 × 50= 40 NThis gives.

a = – 310

But then direction friction of friction we assumedis wrong a = 0friction Ps static and just sufficient to balance forcesfrom equation (ii)

f = 10 N

Q.25 [Ans.

2

112g

, 22mg

]

Sol.

A AB Bf

30° mgsin mgcos

N T

a

Tmgsin

mgcos

N mgcos

for block Amgsin – T = ma ...(1)for block Bmgsin + T –Bmgcos = Ma ...(2)solving (1) & (2)

g –32

·23

g = 2a a =

2g

2

11

T = mgsin – ma = 2mg

–

2

112

mg = 22

mg

Q.26 [Ans. (a) 4g

, (b) 8m5

]

Sol.m1

m2

TT

m1 + m2 = m

Tmax = 32mg15

let m1 > m2m1 g – T = m1 aT – m2 g = m2 a

T =21

21

mmgmm2

...(i)

a =

21

21

mmgm–m

...(ii)

Tmax =

32gm–m15 21

for rimiting condition,

gmmmm2

21

21

= g32

)m–m(15 21

212

21 mm4)mm(1615

(m1 – m2)2 = (m1 + m2)

2 – 4 m1m2

(m1 – m2)2 =

16)mm( 2

21

(m1 – m2)2 =

4)mm( 2

21

substituting the value of (m1 – M1) in equation (ii)

a = 4g

Let maximum value of greater mass be M

Tmax = gm

)M–m(M2 = g

3215

gm

)M–m(M2 = 32

mg15

M = 22m


Q.27 [Ans: a1 = 31g15

m/s2, a2 = 49g15

m/s2, ]

[Sol. Applying Newton's second law in vertical,

N – mg = ma sin37º

N – mg = m( 53

a)

N = m(g + 53

a1)

Also, f = ma cos 37º

N = m( 54

amax)

amax = a1

103

m

1a

53g = m

1a54

a1 = 15g/31 m/s2

Now, for maximum acceleration,

N = m

2a

53g

N = m

2a54

On solving, a2 = 15g/49 m/s2 ]

Q.28 [Ans : 0.54]Sol. m = 4.1 kg

F = 40 Nby figure, vf= 5m/s; v = 0.5m/s

a = tv–v if

= 15.0–5

= 4.5 m/s2

ma = F – mg4.1 × 4.5 = 40 – × 4.1 × 9.8 = 0.54

EXERCISE # 5 PREVIOUS YEAR1. (C)

Tcos Tcos

Tsin Tsin

2 mgmg mg

A

T T

CBTT

For equilibrium in vertical directon for body Bwe have

2 mg = 2T cos

2 mg = 2 (mg) cos[ T = mg (at equilibrium)]

cos = 21

= 45º

Q.2 (D) At equilibrium T = Mg

mg TT

Mg

T

F.B.D. of pulleyT=Mg

F F = (m+M)g1

The resultant force on pulley isF = 22 TF

F = [ 22 M)Mm( ]g

3. (A) The forces acting on the block are shown. Sincethe block is not moving forward for the maximum


force F applied, thereforeF cos 60º = f = N ....(i)[Horizontal Direction]For maximum force F, the frictrional force is thelimiting friction =N]and F sin 60º + mg = N ....(ii)From (i) and (ii)

60º

Fcos60º

FFsin60º

mgf

N

F cos60º = [F sin60º + mg] F =

=

23

321–

21

10332

1

= N20

415

4. (C) JUst before the string is cut by equilibrium of massm, T' = mg ....(i)By equilibrium of mass 2m, T = 2mg + T

....(ii)From (i) and (ii), T = 2mg + mg = 3mg

5. (B)

8. (A) µtan

mg

P

N

force for which f = 0 P0 = mg sin Case I P = mg (sin – µ cos ) P < P0

mg

PfN

P + f = mg sin mg (sin – µcos) + f = mg sin f = µ mgcos

Case II P = mg (sin + µ cos) > P0

mg

PN

f

P = f + mg sinmg (sin + µ cos) = f + mg sin f = µ mg cos ]

Q.9 5Sol. mg (sin + mcos ) = 3 mg (sin – cos )

= 0.5 N = 5

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