Question

Three blocks a b and c of masses 4kg 2kg and 1 kg are in contact on frictionlesss surface. If the force of 14n is applied on the block 4kg, then contact force between a and b is​

Answer 1

Answer :-

$N = 6N$

Given :-

$m_1 = 4kg \\ m_2 = 2kg \\ m_3 = 1 kg \\ F = 14 N$

To find :-

The contact force between a and b is.

Solution:-

All the blocks are in contact with each other.

Friction force is negligible.

Then,

The acceleration of the system is :-

→$F = M a$

→$a = \dfrac{F}{(m_1 + m_2 + m_3)}$

• Put the given values.

→$a = \dfrac{14}{(4+2+1)}$

→$a = \dfrac{14}{7}$

→$a = 2m/s^2$

hence,

Acceleration of the system will be 2m/s².

• The direction of acceleration is toward the applied force.

→$F - N = ma$

→$14 - N = 4 \times 2$

→$14 - N = 8$

→$N = 14 -8$

→$N = 6N$

hence,

The normal force between a and b is 6 N.