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Solution Friction (23)

1)When F=10N applied on 2kg Block on top:

So equation of motion for 2 kg block will be:
F - f = ma
10 - 2(0.2)(10) = 2(a_max)
6 = 2a_max
a_max = 3 m/s².

So the 2 kg block will move with a₁ = 3m/s².

Now since there is no Friction between Ground & 7kg block
So the frition force between 3 kg and 7 kg will drive them
together.

But since the force driving 3 kg block itself f = 4N which
is < f_3/7 = (0.3)(3)(10) = 9N.

Therefore the friction force between 2 kg and 3 kg will
drive the 3 kg and 7 kg blocks together.

So we have.
f = ma
4 = (3+7)(a)
4/10 = a
0.4 m/s² = a = a₂ = a₃.

2)When Force is applied on 3 kg Block:

Here 2 kg and 3 kg block act as system.
f = (2+3)(0.3)(10) = 5*3 = 15N.

Now this friction force of 15N > F applied = 10N so the
Force F = 10N will drive the Whole of System together.
F = (2+3+7)a
10/12 = a
a = 5/6 m/s².
a₁ = a₂ = a₃ = (5/6) m/s².

3)When Force is applied on 7 kg lower Block:
Since no Friction between ground adn 7kg block.so we have
F = (2+3+7)a
10 = 12(a)
5/6 = a

Therefore,
a₁ = a₂ = a₃ = (5/6) m/s².

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