Three blocks of mass 4 kg, 2 kg, 1 kg respectively are in contact on a frictionless table as shown in the figure…

Q: Three blocks of mass 4 kg, 2 kg, 1 kg respectively are in contact on a frictionless table as shown in the figure. If a force of 14 N is applied on the 4 kg block, the contact force between the 4 kg and the 2 kg block will be

Numerical

(a) 2 N

(b) 6 N

(c) 8 N

(d) 14 N

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Ans: (b)
Sol: Acceleration of the system , $\displaystyle a = \frac{14}{(4+2+1)} $

a = 2 m/s2 …(i)

Let contact force between 4 kg & 2 kg block is Fc ,on 2 kg block it is acting towards right & on 4 kg block it is acting towards left.

For 4 kg block ,

$\displaystyle 14 – F_c = 4 a $

$\displaystyle 14 – F_c = 4 \times 2 $

Fc = 6 N