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4 : Block on block systems - Friction

Questions

1.
In the figure shown, if coefficient of friction is μ, then m2 will start moving upwards if:

(a)
m1
> sin θ − μ cos θ
m2
(b)
m1
> sin θ + μ cos θ
m2
(c)
m1
> μ sin θ − cos θ
m2
(d)
m1
> μ sin θ + cos θ
m2
2.
Determine the time in which the smaller block reaches other end of bigger block in the figure

(a)
4s
(b)
8
(c)
2.19 s
(d)
2.13 s
3.
Two blocks are connected over a massless pulley as shown in figure. The mass of block A is 10 kg and the coefficient of kinetic
friction is 0.2. Block A slides down the incline at constant speed. The mass of block B in kg is:

(a)
5.4
(b)
3.3
(c)
4.2
(d)
6.8
4.
In the shown arrangement mass of A = 1 kg, mass of B = 2 kg. Coefficient of friction between A and B = 0.2.

There is no friction between B and ground. The frictional force exerted by A on B equals.
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(a)
2N
(b)
3N
(c)
4N
(d)
5N
5.
A block A with mass 100 kg is resting on another block B of mass 200 kg. As shown in figure a horizontal rope tied to a wall
holds it. The coefficient of friction between A and B is 0.2 while coefficient of friction between B and the ground is 0.3. the
minimum required force F to start moving B will be

(a)
900 N
(b)
100 N
(c)
1100 N
(d)
1200 N
6.
Find the friction force acting on 6 kg block. Ground surface is smooth

(a)
0N
(b)
9N
(c)
12 N
(d)
15 N
7.
In the given figure the coefficient of friction between two blocks is
μ

and all other surfaces are smooth. Find minimum value of F which will prevent slipping

(a)

(m1 + m2 ) m1 g
m2
(b)

(m1 + m2 ) μg
m1
(c)

(m1 + m2 ) g
μ

(d)

(m1 + m2 ) g
μm2
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Answer Key

1. B 2. C 3. B 4. A 5. C 6. C
7. C

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Solutions

1. (B)

m2
will start moving upwards if

m1 g > m2 g sin θ + μm2 g cos θ

m1
⇒ > sin θ + μ cos θ
m2
So, option (b) is correct.
2. (C)

Kinetic friction

fk = μ k N

= 0.3 × 2 × 10 = 6N
Acceleration of upper block
= a1

10 − 6
= = 2m/s2
2

a2 =
acceleration of lower block

6 3
= = m/s2
8 4

3 5
a1,2 = a1 − a2 = 2 − = m/s2
4 4

1
Srel = urel t + arel t2
2

1 5 24 24
⇒3=0×t+ × × t2 ⇒ t2 = ⇒t=√
2 4 5 5

3. (B)T = 2.19s

So, option (c) is correct.

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∵
Block A slides down without acceleration.
So
on block B

T = mg & & ....(1)

on block A

1
T + fk = 10g ( )
2

T + μN = 5g

T + 0.2 (10g cos 30∘ ) = 5g

√3
⇒ T + 0.2 × g × 10 × = 5g
2

⇒ T + √3g = 5g
. . . (2)

By (1) and (2)

m = 5 − √3 = 5 − 1.732 = 3.27

⇒ m = 3.3kg

so (b) is correct.
4. (A)
Block A moves due to friction. Maximum acceleration of A can be

fmax μmg
or or μg = 0.2 × 10 = 2ms−2 .
m m
If both the blocks move together, then combined acceleration of A and B can be

10
= 3.33ms−2 .
3
Since, this is more than the maximum acceleration of A.

Slipping between them will take place and force of friction will be maximum or

μmA g = 2N.

5. (C)

F = fAB + fBG = μAB ma g + μBG (mA + mB ) g

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Step1: Solve assuming both move together

18
a= = 2m/s2
3+6

Step2: From FBD of 6 kg block

f = 6a = 6 × 2 = 12N

Step3: Check

f ≤ fL

fL = μN1 = 0.5 (30) = 15N

Clearly

12N < 15N ⇒

7. (C)

F
a= ;
m1 + m2
, Now taking FBD of only
m1

Horizontally

m1 F
N = m1 a =
m1 + m2

Vertically

μm1 F
μN = m1 g ⇒ = m1 g
m1 + m2

Solving

(m1 + m2 ) g
F=
μ

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