Question

See the attachment below
Q2

Answer 1

Answer:-

a = 1 m/s²

N = 8 N

N' = 5 N.

Given :-

$\mathsf{ F = 10 N, m_1 = 2kg , m_2 = 3kg , m_3 = 5 kg}$

To find:-

The acceleration of the whole system.

And the contact force between :-

1) 2 kg and 3 kg block.

2) 3 kg and 5 kg block.

Solution:-

The acceleration produced in the system is :-

• From Newton 2nd law.

→$\mathsf{ a = \dfrac{F}{(m_1+m_2+m_3)}}$

→$\mathsf{a = \dfrac{10}{(2+3+5)}}$

→$\mathsf{a = \dfrac{10}{10}}$

→$\mathsf{a = 1 m/s^2 }$

hence,

The acceleration produced in the system is 1m/s².

Now,

The Force acting between 1kg and 2kg block.

• Take 1 kg block as system.

The force acting on block 1kg is see in attachment.

→$\mathsf{ F - N = m_1 a}$

→$\mathsf{ 10 - N = 2 \times 1}$

→$\mathsf{10 - N = 2}$

→$\mathsf{10 - 2 = N }$

→$\mathsf{N = 8 N }$

hence,

The contact force between 2kg and 3kg block will be 8N.

Let the contact force between 3kg and 5kg block be N'.

• Take 3 kg block as system.

The force acting on 3kg block is see attachment.

→$\mathsf{N - N' = m_2 a }$

→$\mathsf{8 - N' = 3 \times 1}$

→$\mathsf{8 -3 = N'}$

→$\mathsf{N' = 5}$

hence,

The contact force between 3kg and 5kg block is 5 N.