A block of mass 15 kg is lying on an inclined plane of angle `30^(@)`. In order to make it move upward along the slope with an acceleration 25 `cms^(2)`, a horizontal force of 200 N is required to be applied on it. Then the frictional force on the block i s (g = 9.8 ms 2)

Options

A. 95.95 n
B. 134 n
C. 99.70 n
D. 90 n

Answer

C

Explanation

From the figure if the block is moving upward along the slope with an acceleration `a_(x )` them `Sigma F_(x ) ne 0 ` but `Sigma F_(x ) = ma_(x )`
`200 xx 0.866-f -15 xx 9.8 xx (1)/(2) =15 xx 0 .25`
`173.2 -f -73.5 = 3.75`
`:. , 95 . 95 N=f`