Multiple choice questions on mechanics

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Topic :
Mohammed
Asif Ph : 9391326657, 64606657

Multiple choice questions (only one is correct)
1. A body of specific gravity 6, weighs 0.9kg when placed in one pan (say pan A) and 1.6kg when
placed on the other pan (pan B) of a false balance. The beam is horizontal when both the
pans are empty. Now if the body is suspended form pan A and fully immersed in water, it
will weigh
a) 0.5kg b) 0.6kg c) 0.75kg d) 0.8kg

2. The pulley has mass M >m. String is massless. The above system is released from rest from the
position shown. Then

a) Body (1) will slowly come down, till equilibrium is attained at level AA’ shown.
b) The system will perform oscillations with equilibrium position at level AA’ and amplitude
0.5m
c) The response of the system depends on whether pulley-string interface has friction or not.
d) The system will continue to be in the same initial position.

3. In the position shown above Let T1 be the tension in the left part of the string, T2 = Tension in
the right part, N = Normal reaction on block (2) by the resting surface. Then

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a) T1 = T2 = 0 and N = mg b) T1 =mg, T2 = 0, N =mg
mg mg
c) T1 = T2 = mg and N = 0 d) T1 = T2 = and N=
2 2

4. A body of mass m was slowly hauled up the hill and down the hill onto the other side by a
→
force F which at each point was directed along a tangent to the trajectory. The work
performed by this force, if the coefficient of friction is µ1 uphill and µ 2 dowbhill, is

a) mg ( µ1 1 + µ 2  2 ) b) mg ( 2h + µ1 1 + µ 2  2 )
c) mg ( h + µ1 1 + µ 2  2 ) d) mg ( µ1 1 − µ 2  2 )

5. The bob of a pendulum is taken to position A and given an initial velocity u in the direction
shown, in the following cases.

If the minimum value of u so that the pendulum reaches position OB in case (i) and position
OC in case (ii) are u1 in case (i) & u2 in case (ii), then
u2
u1
(
is g = 10 ms −2 )
a) (3 + 2 ) b) (2 + 3 ) c) (2 + 3 ) d) (3 2 −2 )


6. A conveyor belt carrying powdery material is at an angle 370 to the horizontal and moves at a
constant speed of 1ms-1 as shown. Through a small hole in the belt, the powdery material
drops down at a constant rate of 1kg per second. What is the force to be applied on the belt
along the direction of its motion so as to maintain its constant speed of 1ms-1?

a) -1 N b) Zero c) +1 N d) None of the above

7. Small body A on a hemispherical body B which is on a wedge C which is on a smooth
horizontal surface. System is released from rest from the position shown when
α = 37 0 ,  B = 10cm,  = 5cm .
When α = 53 ,  C is 4.5cm  B at that instant is (neglect friction)
0

a) 9 cm b) 10 cm c) 10.5 cm d) 11 cm

8. A pendulum consists of a mass m attached to the end of a light string 0.5m long. It can
oscillate in the vertical plane. If it is let go in the horizontal position and has an inelastic
collision with the floor, e = 0.5, the rebound velocity is (ms -1) (the angle turned is π / 6 ) (g =
10ms-2)

5 5 35 45
a) b) c) d)
2 4 4 4

9. The track is in the vertical plane. The track is rough with friction coefficient µ . A particle at
A is allowed to slide down and goes upto B and returns. Heights of A and B are 0.2m and
0.1m respectively, as shown. The maximum possible value of µ is


1 1 1 1
a) b) c) d)
4 7 6 5

10. A particle is projected form a point on a horizontal floor. After it has three collisions with the
 1012 
floor, it is found that the ratio of maximum height to minimum peaks reached by it is  36 
2 
 
. The coefficient of restitution is
a) .5 b) 0.64 c) 0.8 d) 0.9

b
11. The potential energy function along the positive x axis is given by U ( x ) = − ax + , a, b are
x
constants. If it is known that the system has only one stable equilibrium configuration, the
possible values of a and b are
a) a = 1, b = 2 b) a = 1, b = -2 c) a = -1, b = 2 d) a = -1, b = -2

12. The acceleration of the 1kg block immediately after the string is cut is (g = 10ms-2).

a) 4ms-2 b) 4.1ms-2 c) 16ms-2 d) 40ms-2

13. A circular pan with its side wall inclined inward at 530 as shown is rotating about its central
vertical axis with a constant angular velocity. A ball placed at the edge rotates along with it
as shown. If the ball exerts a force of 22.5 N on the side wall and 23.5N on the bottom
surface, the mass of the ball is


a) 1kg b) 2kg c) 3kg 4kg

14. Acceleration of 10kg block, when system released from rest, is

a) 0.5ms-2 b) 1 ms-2 c) 1.25 ms-2 d) 1.66ms-2

15. Case(A)

Case (B)

The reaction force between the 5kg and 6kg block in case A and case B will be
a) equal and non zero b) unequal with case A being more
c) unequal with case being more d) equal and zero

For Question No. 16 to 20:
Each question consider of two statements: one is Assertion (A) and the other is Reason(R).
You are to examine these two statements and select the answer using the code given below.
a) Both A and R individually correct, but R is the correct explanation of A
b) Both A and R individually correct, but R is the not correct explanation of A
c) A is true but R is false d) A is false but R is true
16. Assertion (A): If a block is released form rest, when reaches the bottom point of the wedge, its
speed is same irrespective whether the wedge is fixed or the wedge is fee to
move.
Reason(R): Mechanical energy is conserved in both cases.

17. Assertion (A): A body is at rest on floor. You lift it vertically up and bring it to rest at a point
h
above the ground. The work done by you is zero.
Reason(R): Any non-zero work done by a force on a body results in change in kinetic energy
of
the body.


18. Assertion (A): The negative of the work done by the conservative internal forces on s system
equals to change in its potential energy.
Reason(R): Work energy theorem.

19. Assertion (A): In a tug of war that team wins which applies more tension force on string then
their opponents.
Reason(R): The winning team must be having stronger players.

20. Assertion (A): If a block starts moving at t = 0 with an acceleration ‘a’ then its kinetic energy
at
time t with respect to two non-inertial frames which have same acceleration in a
direction is not always same.
Reason(R): The work energy theorem ∆ K − ∆ W is not for non-inertial frame.

Match the following Questions 21 to 25
21. Two columns are given in each question. Match the elements of Column-I with Column-II.
Column-I Column-II
i) Friction force p) Contact force
ii) Normal reaction q) Electromagnetic force
iii) Tension in a string r) Gravitational force
iv) Force between two charges of mass m s) Nuclear force
a) (i-p,q), (ii-p,q), (iii-q,s), (iv-q,r) b) (i-p,q), (ii-p,q), (iii-q), (iv-q,r)
c) (i-p,q), (ii-p), (iii-q,s), (iv-q) d) (i-q), (ii-p), (iii-q,s), (iv-p,q,r)

22. In Column-I there are some motions of a body and Column-II contains the list of concepts
that can be used for the analysis of these motions
Column-I Column-II
i) A body moving in a vertical circle p) Conservation of energy
ii) A body moving in a horizontal circle q) Conservation of momentum
iii) A body dropped form a height on a block attached
to the top of a vertical spring r) Centripetal force
iv) Rocket propulsion s) Centrifugal force
a) (i-p,q,r), (ii-q,r,s), (iii-p,q,s), (iv-p) b) (i-p,s), (ii-p,r), (iii-q), (iv-p,q,r)
c) (i-p,r), (ii-q,s), (iii-p), (iv-p,q) d) (i-p,r,s), (ii-r,s), (iii-p,q), (iv-q)

23. When two bodies collide they come in contact at t = t 1 and loses contact at t = t2. This (t2-t1) is a
very small time interval. Consider this small time interval and match the following.
i) Elastic collision p) Kinetic energy decreases and potential
energy increases and then potential
energy decreases and kinetic energy increases
ii) Inelastic collision q) Kinetic energy + potential energy is
conserved
iii) Perfectly inelastic collision r) Momentum is conserved
iv)Oblique elastic collision s) Kinetic energy decreases and potential energy
increases and then situation remains same


a) (i-p,q,r), (ii-q,r,s), (iii-p,q,s), (iv-p) b) (i-p,s), (ii-p,r), (iii-q), (iv-p,q,r)
c) (i-p,r), (ii-q,s), (iii-p), (iv-p,q) d) (i-p,r,s), (ii-r,s), (iii-p,q), (iv-q)

24. For each of the following four cases of Column-I match the range of force F in column B so
that the block m is not slipping on the surface with which it has contact.
For all cases m = 2kg; µ = 0.2 Take g = 10m/s2
Column-I Column-II

p) (48N, 80N)

 500 500 
 N, N
 19 11 

q)

 220 380 
r)  N, N
 19 11 

 220 380 
 N, N
 23 17 
s)
a) (i-p), (ii-q), (iii-q,r), (iv-p,q,r,s) b) (i-p), (ii-q), (iii-r,s), (iv-q,r)
c) (i-q), (ii-p,s), (iii-q,s), (iv-r,s) d) (i-q,r), (ii-p,q), (iii-r), (iv-p,q,r)

25. A body is moving in a vertical circle of radius R, considering motion from top point of circle
bottom most point of circle, match the following.
Column-I Column-II
i) Tangential accelertaion p) Always increases
ii) Centripetal acceleration q) Always decreases
iii) Angular velocity r) First increases then decreases
iv) Potential energy s) Variable depending on angle made with
vertical by string
t) Variable independent of the made angle with
vertical by string
a) (i-r,s), (ii-q,r), (iii-p,t), (iv-p,q) b) (i-q,s), (ii-r,s), (iii-p,s), (iv-q,r)


c) (i-p,s), (ii-q,r), (iii-q,s), (iv-p,r) d) (i-r,s), (ii-q,s), (iii-p,s), (iv-p,s)

Write the final answer to each question in this section in the column provided.
26. The block ABCDE of mass 5m has BC part spherical, of radius 1m, is on frictionless
horizontal surface. BC is quarter circle. A small mass m is released at B and slides down.
How far away form D does it hit the floor? (g = 10ms-2)

27. The kinetic energy of a particle of mass 1kg moving along a circle of radius 1m depends on
time t as K = t4. Fid the force acting on the particle as function of t.

28. Spring is already in a compressed potion with initial compression = 5cm. The 10kg block is
allowed to fall. Determine the maximum compression of the spring before the block
rebounds. (g = 10ms-2).

29. A long plank of mass M = 8kg, length  = 1m rests on a horizontal surface. Coefficient of
friction µ1 = 0.2 . A small block, mass m = 2kg rests on the right extreme and of M, on the
rough top surface. At t = 0, a force F= 25.5N is applied on M towards the right. (see figure).
The block m does not slide on M. At t = 3s, force F is increased to 31N. The block m falls off
M’s surface at t =7s. Determine the coefficient of friction between m and M.

30. A particle is released on the smooth inside wall of a cylindrical tank at A with a velocity u
which makes an angle α with the horizontal tangent. When the particle reaches a point B, a
distance h below A, determine the angle β (as an inverse function of cosine) made by its
velocity with the horizontal tangent at B.


Passage-I (31 to 33):
Two blocks of masses 10kg and 5kg are placed on a rough horizontal floor as shoen in figure.
The strings and pulley are light and pulley is frectionless. The coefficient of friction between
10kg block and surface is 0.3 while that between 5kg block and surface is 0.2. A time varying
horizontal force. P=5t Newton (t is in sec) in applied on 5kg block as known. [Take g=10ms-2]

31. The motion of block starts at t = t0, then t0 is
a) 14s b) 8s c) 9s d) 12s

t0
32. The friction force between 10kg block and surface at t = is in between
2
a) zero and 10N b) 10N and 35N c) 12.5N and 17.5N d) 12.5N certain value

33. The acceleration of 5kg block at t = 2t0 is
14 14 −2
a) 12 ms-2 b) ms −2 c) ms d) 2 ms-2
5 9

Passage-II (34 to 36):
A block of mass 4kg is pressed against a rough wall by two perpendicular horizontal force F1
and F2 as coefficient of static friction between the block and floor is 0.6 and that of kinetic
friction is 0.5. [Take g=10ms-2]

34. For F1 = 300N and F2 = 100N, find the direction and magnitude of friction force acting on the
block.
a) 180N, vertically upwards b) 40N, vertically upwards


−1  2 
c) 107.7N, making an angle of tan   with the horizontal in upward direction.
5
−1  2 
d) 91.6N, making an angle of tan   with the horizontal in upward direction.
5

35. For F1=150N and F2=100N, find the direction and magnitude of friction force action on block.
−1  2 
a) 90N, making an angle of tan   with the horizontal in upwards direction
5
−1  2 
b) 75N, making an angle of tan   with the horizontal in upwards direction
5
−1  2 
c) 170.7N, making an angle of tan   with the horizontal in upwards direction
5
d) Zero

36. For data of Question No.35, find the magnitude of acceleration of block.
a) Zero b) 22.5 ms-2 c) 26.925 ms-2 d) 8.175 ms-2

37. System shown in figure is in equilibrium and at rest. The spring and string are massless, now
the string is cut. The acceleration of mass 2m and m just after the string is cut will be

g g
a) upwards, g downwards b) g upwards, downwards
2 2
c) g upwards, 2g downwards c) 2g upwards, g downwards


KEY
1) c 2) d 3) c 4) a 5) d
6) b 7) c 8) c 9) b 10) b
11) c 12) d 13) a 14) d 15) c
16) d 17) d 18) c 19) d 20) c
21) b 22) d 23) c 24) a 25) d
26) 24
m
27)
β 2 + t6 N 28) 5
2
( )
17 − 1 cm
29) µ 2 = 0.1 30) 



5 −1  u cos α 
cos
 2 gh 
 1+ 2 
 u 
31) a 32) c 33) c 34) c 35) b
36) d 37) a

Solutions
1. i)

m1 1 = m2  2 ……….(1)

ii)

m1 1 + M 2 = m2  2 + 0.9 2 ⇒ M 1 = 0.9 2 …………(2)

iii)

m1 1 + 1.6 1 = m2  2 + M 2
⇒ 1.6 1 = M 2 ………….(3


iv)

1
m1 1 +  1 = m2  2 + x 2 ⇒  1 = x 2 ⇒ x =
2
From (2) M 1 = 0.9 2
 0.9 3
⇒ 1.2 1 = 0.9 2 ⇒ 1 = = = 0.75
 2 1.2 4
x = 0.75kg

2. (d) Self explanatory [acc = 0, u = 0 ∴ at rest]

3. (c) Self explanatory [acc = 0, u = 0 ∴ at rest]

4. Uphill: F = mg sin θ + µ1 mg cos θ
F.ds = mg ds sin θ + µ1 mg ds cos θ
∫ F .ds = mg ∫ dy + µ 1 mg ∫ dx
= mg .h + µ1 mg  1 ………(1)
Downhill:
− F = mg sin θ − µ 2 mg cosθ
F .ds = Fds ⇒ ∫ F .ds = − mgh + µ 2 mg  2 ……….(2)
∴ W = (1) + ( 2 ) = mg ( µ1 1 + µ 2  2 )

5. Case (i)

2
muC
TA ≥ 0 ⇒ = mg cos 45 ⇒ u12 > gR …………..(1)
R
Case (ii)


2
mvC
TC ≥ 0 ⇒ ≥ mg ⇒ vC > gR
2

R
1 1 2  1 
mu 2 = mvC + mgR1 − 
2 2  2
⇒ vC = u 2 − 2 gR + 2 gR ⇒ u 2 − 2 gR + 2 gR > gR
2

(
⇒ u 2 = 3 − 2 gR
2
) ………….(2)

⇒
u2
2
=
( 2) = 3 − 2 = 3 2 − 2 ⇒ u2 = (3 2 −2 )
u2
1 (1) 1 / 2 u1

dm
6. Freaction = .vrel . But vrel = 0 .
dt

7. No external force horizontally,
∴ xCM does not shift
Taking →
x A / B = 0.3( sin 53 − sin 37 ) = 0.06 m
x B / C = unknown
xC = given = −0.005m
∴m A ( x A / B + x B / C + xC ) + mB ( x B / C + xC ) + mC ( xC ) = 0
⇒ 1( 0.06 + x B / C − 0005) + 2( x B / C − 0.005) + 3( − 0.005) = 0
⇒ x B / C = 0.01 ⇒ x B = x B / C + xC
= 0.01 − 0.005 = 0.005m.
⇒  B = 0.10 + 0.05 = 0.105m = 10.5 cm

8. Before collision:
Using loss of P.E = Gain of KE

3
Components u y = − 5.
2


5
ux = −
2
After collision
15 −1
v y = −e.u y = + ms
4
5 35 −1
vx = u x = − ∴v = v y + vx =
2 2
ms
2 4

9. Work done by friction, wf.
We know wf has to be minimum

= µ mg ( 0.4 + 0.3) = 0.7, µ mg .
But wf = loss of P.E = 0.1mg
1
∴ 1mg ≥ 0.7 µ mg ⇒ µ <
7

10.

^ ^
u = u sin θ j + u cos θ i
^ ^
u1 = eu sin θ j + u cosθ i
^ ^
u 2 = e 2 u sin θ j + u cosθ i
^ ^
u 3 = e 3u sin θ j + u cosθ i
e 2 sin 2 θ e 6 u 2 sin 2 θ
H= H3 = = e6 H
2g 2g
6
H 1 1012 2 36  2 6 
∴ = 6 = 36 ⇒ e 6 = 12 =  2 
 10 
H3 e 2 10  
6
2 64
⇒e= 2 = = 0.64
10 100


dU ( x ) b
11. = −a + ( −) 2
dx x
dU − ax 2 − b
= 0 ⇒ equillibrium ⇒ =0
dx x2
b
⇒ x ≠ 0 and x 2 = −
a
⇒ b and a are of opposite signs.
⇒ options (b) or (c) possible
For stable equation.
d 2U
should be +ve
dx 2
b b
⇒ ( − )( − 2 ). 3 = 3 > 0 ⇒ ( for x > 0) b > 0
x x

12. T’ =70N, T = T’ -30N=40N
On cutting thread,
40
A1kg = = 40ms −1
1
All other’s acc = 0

13. N1 = N 2 sin 37 + mg ……..(1)
N 2 cos 37 = mω R2
……...(2)

∴ N1 = 23.5
N 2 = 22.5
⇒ (1) ⇒ 23.5 = 22.5 × 0.6mg
⇒ mg = 23.5 − 13.5 = 10 N ⇒ m = 1kg


14.

f static = 20 N
20
∴A= =2
10
25 − f 25 − 20
∴a = =
5 5
a < A ⇒ not possible
Hence no relative motion
25
⇒ A=a= = 1.66 ms − 2
10 + 5

15. Case(A)

1 − R1 = 40a ⇒ R1 = 1 − 40a
Case (B)

1 − R2 = 15a ⇒ R2 = 1 − 15a
∴ R2 > R1

16. R is clearly true, but A is false. Initial energy is P.E. If wedge is free to move, it will have KE ⇒
KE of block will be less than that if wedgeis fixed.

17. You did positive work and gave positive energy. Gravity did equal negative work and made net
∆ KE zero.

18. Self explanatory.


19. Tension in rope remains same through out but it is the frictional force provided by the ground
that helps a team to win.

20. Assertion is right because only the charge in kinetic energy in both the frames are same
(including work of pseudo force) but kf is dependent on the ki as
kf – ki = ∆ k
and ki depends on the velocities of these reference frames at time t =0 reason is wrong
so assertion is right and reason is wrong

21. (b) self explanatory

22. (d) self explanatory

23. (c) self explanatory

24. (a) 1st case

20 cos 37 = f ≤ uN (for no motion)
16 ≤ 0.2( 20 + F + 20 sin 37 )
F ≥ 48 N
F ∈ ( 48, ∞ ) ………….(1)
nd
II case

mg − F sin 37 = f ≤ µN (to prevent downward slipping)
3F
20 − ≤ 0.2( F cos 37 )
5
 4 3
F  +  ≥ 20
 25 5 
500
F≥ N
19
F sin 37 − mg = f ≤ µN (to prevent upward slipping)


3F  4
− 20 ≤ 0.2 F 
5  5
3 4 
F  −  ≤ 20
 5 25 
500  500 500 
F≤ N so F ∈  N, N ……..(2)
11  19 11 
IIIrd case

mg sin 37 − F sin 37 = f ≤ µN (for no downward slipping)
3  4 
12 − F   ≤ 0.2 F   + 16 
 5 
5    
 4 3 16
F  +  ≥ 12 −
 25 5  5
44 25 220
F≥ × = N
5 19 19
− mg sin 37 + F sin 37 = f ≤ µN (for no upward slipping)
3F  4 
− 12 ≤ 0.2 F   + 16 
 5 
5    
 3 4  76 76 25 380
F −  ≤ ⇒F≤ × = N
 5 25  5 5 11 11
 220 380 
So, F ∈  N,  ……………(3)
 19 11 
IVth Case

f = mg sin 37 − F cos 37 ≤ µN (for no downward sliding with respect to wedge)
4F  3
12 − ≤ 0.216 + F 
5  5
4 3  16 220
F  +  ≥ 12 − ⇒ F ≥ N
 5 25  5 23
F cos 37 − mg sin 37 = f ≤ µN (for no upward sliding with respect to wedge)


4  3F 
F − 12 ≤ 0.216 + 
5  5 
380
F≤
17
 220 380 
F ∈ N, N ……….(4)
 19 17 

25. (d) Self explanatory

26. When m leaves C, let its velocity (horizontal) be u.
u
Then block 5m will have velocity to left.
5
∴ Energy equation:
5 gR 50
⇒u = = ms −1
3 3
∴ t = time to reach ground = 2h = 2 ×1 = 1 s
g 10 5
∴ Distance of hitting point from D =
u 6u 6 50 1 24
= ×t + u×t = t= . = m
5 5 5 3 5 5

1 1
27. K = mv 2 = v 2 = t 4 ………..(1)
2 2
2
mv 1.v 2
∴ Fn = = = v 2 = 2t 4
R 1
dv
v 2 = 2t 4 v = 2 t2 = 2 2t
dt
dv dv
Ft = m = 1. = 2 2 t
dt dt
∴ Resultant F = Ft 2 + Fn2 = 2 t 2 + t 6

28. Initial energy:
1 1
P.E spring = kx 2 = × 4000 × ( 0.05) = 5 J
2

2 2
P.E mass = mgh = 100 × 0.2 = 20 J
∴ Total Einitial = 25 J
Final energy
P.E spring = k ( x + x 2 ) = × 4000 × ( 0.05 + x')
1 1 2

2 2


× 4000 × ( 0.0025 + x' 2 +0.1x')5 + 2000 x '2 +200 x '
1
2
P.E mass = − mgx' = −100 x '
∴ 2000 x '2 +100 x − 20 = 0

⇒ x' =
5
200
( )
17 − 1 m =
5
2
( )
17 − 1 cm

29. From t = 0 to t = 3sec

f2 < f2 max
so both blocks are moving together and then
F – f1 max = (m + M) a ………(1)
f1 max =0.2 x 10 x 10 = 20N
So 25.5 – 20 = 10a
⇒ a = 0.55 m / s −1 ……….(2)
1
Displacement S1 = × 0.55 × 3
2

2
=2.475m ……….(3)
Velocity = 0.55 x 3
= 1.65m/sec ……….(4)
From t = 3 to t = 7 sec
f2 =f2 max = 20 µ 2
31 − 20 − 20 µ 2 = 8 am
20 µ 2 = 2am
1
1 = ( aM − am ) 4 2
2
1
⇒ a M − am =
8
After solving these equation.
9
a M = , am = 1, µ 2 = 0.1
8
30. Initial velocity u can be split into tangential component
u cos α and vertical component u sin α
∴ vz at B can be attained by,
v z = u 2 sin 2 α + 2 gh
2

mu 2 cos 2 α
Normal reaction N =
r
∴ tangential component at B = u cos α


∴ velocity at B = v 2 + u 2 cos 2 α = u 2 + 2 gh
z
u cos α
cos −1
∴ ∠β is given by 2 gh
1+ 2
u

31 to 33.
From constraint theory we can relate the acceleration 10kg and 5kg blocks.

Limitng friction force between 5kg blocks and surface is, f L1 = 0.2 × 5 × 10 = 10 N
Limiting friction force between 10kg block and surface is, f L2 = 0.3 ×10 ×10 = 30 N
For 5kg block, P-2T - f1 = 0 [in equilibrium, i.e., when blocks are not moving]
For no motion of blocks, f1 > f L1 and f 2 ≤ f L2
So, P − 2T ≤ f L1 and T ≤ f L2
⇒ P ≤ f L1 + 2 f L2
So, for motion to take place, P ≥ f L1 + 2 f L2
⇒ 5t 0 = f L1 + 2 f L2 = 10 + 2 × 30
⇒ t 0 = 14 s
t0
At t = = 7 s , the equation for 5kg and 10kg blocks are
2
35 − 2T − f1 = 0 and T − f 2 = 0
⇒ 35 = f1 + 2 f 2
And we know at t = 7s both the blocks are at rest so
f1 ≤ f L1 and f 2 ≤ f L2
Solving above equation we get, 0 ≤ f1 ≤ 10 N ;
12.5 N ≤ f 2 ≤ 17.5 N
And 12.5 ≤ T ≤ 17.5 N
At t = 2t0 = 28s, equation for blocks are
140 – 2T – 10 = 5a and T – 30 = 10 x 2a
14
⇒ a= ms −2
9

34. The forces acting on the block are F1, F2, mg, normal contact force and frictional force. Here
fractional force won’t act along vertical direction as the component of resultant force along the
surface acting on body is not along vertical direction and direction of the friction force is either
opposite to the motion of block (direction of acceleration of block) if it is moving or not moving.


So, f L = µN1 = 0.6 × 300 = 180 N
−1  2 
Resultant of 4g and F2 is 107.7N making an angle of tan   with the horizontal
5
As force applied along the surface is > f L , so the block doesn’t move and friction is static in
nature.
2
f = 107.7 N making an angle of tan −1   with the horizontal in upward direction.
5

35. For F1 = 150 N , f L = 0.6 × 150 = 90 N
As component of resultant force along the surface is 107.7N and is greater than f L , so kinetic
friction comes into existence, i.e., friction force acquires the value f = µ k N1 = 0.5 × 150 = 75 N .
Its direction is opposite to component of resultant force along the surface.

107.7 − 75
36. Acceleration of block = = 8.175 ms −2
4

37. Initially under equilibrium of mass m: T = mg
Now, the string is cut. Therefore, T = mg force is decreased on mass m upwards and
downwards on mass 2m.
mg
∴ am = = g (downwards)
m
mg g
And a2 m = = (upwards)
2m 2


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