For the situation shown in Fig the block is sationary w.r.t.inclined fixed in an elevator the alevator is having an acceleration of `sqrt(5)a_(0)` whose components are shown in the figure the surface is rought and coefficient of static friction between the inclined and block is `mu` Determine the magnitude of force excerted by inclined on the block (Take` a_(0) = g//2 and theta = 37^(@) , mu_(s) = 0.2)`

Options

A. `(mg)/(10)`
B. `(9 mg)/(25)`
C. `(3mg)/(25) xx sqrt(41)`
D. `(sqrt(13)mg)/(2)`

Answer

d

Explanation

The FBD of block from lift is as shown in figure from given data as `m(g +a_(0)) sin theta gt 2 ma_(0) cos theta`

so friction force acts upwards
`f = m(g +a_(0)) sin theta - 2 ma_(0) cos theta`
`= (9g)/(10) - (4 mg)/(5) = (mg)/(10)`
`N = m(g +a_(0)) sin theta + 2 ma_(0) sin theta = (9mg)/(5)`
`As f_(L) = mu_(s) N= (18 mg)/(50) = (9mg)/(25) gt f` so static friction.
Reaction solution
`R = sqrt(f^(2) +N^(2)) = (mg)/(5) sqrt((1)/(4)+9^(2)) = (mgsqrt(13))/(2)`
Alternative solution:
Net force`barF - mg hati = m(a_(0) hati - 2a_(0)hati)`
`rArr barF = m(-2a_(0) hati + (a_(0)+ g)hatj) = m(-g hati + (3g)/(2)hatj)`
`rArr F = msqrt(g^(2) +((3g)/(2))^(2)) = (sqrt(13)mg)/(2)`