Question

A block is placed on an inclined plane. The block is moving towards right horizontal with an acceleration `a_(0)=g`. The length of the inclined plane (AC) is equal to 1 m. Whole the situation are shown in the figure. Assume that all the surfaces are frictionless. The time taken by the block to reach from C to A is (Take,`g=10m//s^(2)`)

A. 0.74 s
B. 0.9 s
C. 0.52 s
D. 1.24 s

Answer 1

Correct Answer - A
The forces on smaller block is given as

For the motion of the block along the incline plane in upward direction.
Net force on the block = mass `xx` acceleration of the block
`implies " "mg cos 30^(@)-mg sin 30^(@)=ma" " [because a_(0)=g]`
`implies " "a=((sqrt(3)-1)/(2))g=3.66 m//s^(2)`
Now, from equation of motion s = `(1)/(2)at^(2)`
`implies" "t=sqrt((2s)/(a))=sqrt((2xx1)/(3.66))=0.74 s`