Question

The contact force between $2\text{kg}$ and $3\text{kg}$ block placed on an inclined plane as shown in the figure will be

• A. $3\text{N}$
• B. $6\text{N}$
• C. $12\text{N}$
• D. $18\text{N}$

Answer 1

The correct option is C $12\text{N}$

Considering both the block as a $5\text{kg}$ system because both will move the same direction.

Using Newton’s second law: $\begin{array}{}N=5g\cos 37°& \text{(i)}\end{array}$
$\begin{array}{}5g\sin 37°-20=ma=5a& \text{(ii)}\end{array}$
$\Rightarrow (5\times 10\times \frac{3}{5})-20=5a$ $\Rightarrow a=2{\text{m/s}}^2$
(Down the inclined plane)

Let the contact force between $2\text{kg}$ and $3\text{kg}$ block be $N_1$. F.B.D. of $3\text{kg}$ block

From $F_{\text{net}}=ma$
$\Rightarrow 3g\sin 37°-N_1=3\times 2$
$\Rightarrow 18-N_1=6$
$\Rightarrow N_1=12\text{N}$