First slide

Making equation on system

Question

In the arrangement shown, the wedge W is moving to the right with constant acceleration ‘a’ and the 2kg block remains stationary relative to the wedge. All surfaces are frictionless. Then the normal reaction of the wedge on the block is $(g = 10 m / s^{2})$

Easy

A

40 N
B

$40 \sqrt{3} N$
C

$20 \sqrt{3} N$
D

$10 \sqrt{3} N$

Solution

(1)
Vertical component of acceleration of the block is zero. So net vertical force acting on the block must be zero.
$\begin{array}{l} ∴ N \cos 60^{0} - m g = 0 \\ \Rightarrow N = \frac{2 \times 10}{\cos 60^{0}} N = 40 N . \end{array}$

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