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)

A block of mass 15 kg is lying on an inclined plane of angle 30^(@). I

1.	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)
Answer» 95.95 n 134 n 99.70 n 90 n Solution : Fromthe figureif theblockis movingupwardalongthe slopewith ANACCELERATION`a_(x )`them `Sigma F_(x ) ne0 ` but `Sigma F_(x )= ma_(x )` `200 xx0.866-f -15 xx 9.8 xx (1)/(2) =15 xx 0 .25` `173.2 -f -73.5 = 3.75` `:. ,95 . 95 N=f`