Question

Three blocks of masses 1 kg, 2kg and 3 kg are placed in contact with each other on a horizontal frictionless surface as shown in figure. A force of 12 N is applied as shown in figure.
Calculate (i) the acceleration of the system of these three blocks, (ii) the contact force acting on 2 kg block by first block of 1 kg and (iii) the contact force on 3 kg block.

Answer 1

Answer: $a = 2 m/s^{2}$, $F_{12} = 6 N$ and $F_{3} = 6 N$.

Explanation:

Given that,

Mass of block 1 = 1 kg

Mass of block 2 = 2 kg

Mass of block 3 = 3 kg

Force = 12 N

(I). The acceleration of the system of these three blocks is

$a = \dfrac{F}{m_{1}+m_{2}+m_{3}}$

$a = \dfrac{12N}{6 kg}$

$a = 2 m/s^{2}$

(II). The contact force acting on 2 kg block by first block of 1 kg is

$F_{12} = a(m_{1}m_{2})$

$F_{12} = 2\m/s^{2}\times3 kg$

$F_{12} = 6 N$

(III). The contact force on 3kg block is

$F_{3} = 6 N$

Hence, this is the required solution.