(i) a 1 = 3 m/s 2 , a 2 = a 3 = 0.4 m/s 2 , (ii) a 1 = a 2 = a 3 = m/s 2 , (iii) same as

Sol. (i) Assuming there is no slipping anywhere and the common acceleration of the three blocks be a

writing the force equation F = ma

10 = 12a or a = .

Now f max. between 2 kg & 3kg is

1 N = 0.2 × 20 = 4N.

For 2 kg block F.B.D. will be 2 kg

F – f = ma = 2 ×

10 – f =

10 – = f

= f > f max

There will be slipping between 2 kg and 3 kg block.

Now considering the slipping the new equation would be

F – f max = ma 1

10 – 4 = 2a 1 or a 1 = 3 ms –2

Now let’s take 3 kg & 7 kg as system and writing the force equation.

f max = 10 a 0 .

or 4 = 10 a 0

a 0 = 0.4 ms –2

To check the required friction between 3 kg and 7 kg block.

F on 7 kg block 7 kg F = 7 × 0.4 = 2.8 N

f max between 7 kg and 3 kg = 0.3 N 2 = 0.3 × 50 = 15 N

2.8 < f max

hence there is no slipping between the two blocks

(ii) Now if force is applied on 3 kg block

Assuming there is no slipping anywhere and the common acceleration of the three blocks is

ms –2

Now, if all system is going with common acceleration ms –2

for a = ms –2 required friction force between 2 kg and 3 kg block = m 1 a = 2 × = N < f max

so there is no slipping

Same for 7 kg block, required friction is = m 3 a = 7 × = N< f max

so there is no slipping

a 1 = a 2 = a 3 = ms –2

(iii) same as