Question

8.
A block of mass 1 kg is at rest relative to a smooth wedge being moved
leftwards with constant acceleration 5m/s2. Let N be the normal reaction
between the block and the wedge. Then N and tano are :
a = 5
(A) N = 5/5 N and tan 0 = =
(B) N = 15 N and tan 0 = -
(C) N = 5V5 N and tan 0 = 2
(D) N = 15 N and tan 0 = 2​

Answer 1

Explanation:

Right answer is option b

Answer 2

Given :-

Mass of block = m = 1 Kg

Acceleration = a = 5 ms-²

In horizontal direction,

NSinØ = ma ------(1)

In vertical direction

NCosØ = mg -----(2)

(1)/(2)

SinØ/CosØ = ma/mg

tanØ = a/g

tanØ = 1/2

|N| = √(ma)²+(mg)²

|N| = 5√5 N

Here, the wedge is moving but the block is at rest relative to wedge. Hence, its frictional force will act in direction of inclination of the wedge.