A block is kept on a frictionless inclined surface with angle of inclination 'α'. The incline is given an acceleration a to keep the block stationary. Then a is equal to :

Question

A block is kept on a frictionless inclined surface with angle of inclination 'α'. The incline is given an acceleration a to keep the block stationary. Then a is equal to : (A) g/ tan α (B) g cosec α (C) g (D) g tan α

Tardigrade Admin · Accepted Answer

In the frame of wedge, the force diagram of block is shown in figure. From free body diagram of wedge. For block to remain stationary, ma cos α = mg sin α a=g tan α

Q. A block is kept on a frictionless inclined surface with angle of inclination ' $α$ '. The incline is given an acceleration a to keep the block stationary. Then a is equal to :

$g / t an α$

$g cosec α$

$g$

$g t an α$

In the frame of wedge, the force diagram of block is shown in figure. From free body diagram of wedge.
For block to remain stationary,
$ma cos α = m g s in α$
$a = g t an α$

Q. A block is kept on a frictionless inclined surface with angle of inclination 'α'. The incline is given an acceleration a to keep the block stationary. Then a is equal to :

g/tanα

gcosecα

g

gtanα

In the frame of wedge, the force diagram of block is shown in figure. From free body diagram of wedge. For block to remain stationary, macosα=mgsinα a=gtanα

Q. A block is kept on a frictionless inclined surface with angle of inclination ' $α$ '. The incline is given an acceleration a to keep the block stationary. Then a is equal to :

$g / t an α$

$g cosec α$

$g$

$g t an α$

In the frame of wedge, the force diagram of block is shown in figure. From free body diagram of wedge.
For block to remain stationary,
$ma cos α = m g s in α$
$a = g t an α$