A block of mass 20 kg is lying on an inclined plane of angle 30^(@). In order to make it move upward along the slope with an acceleration of 25 cm/s^(2), a horizontal force of 400 N is required to be applied on it. (i) Frictional force on the block is .......N. (ii) Co-efficient of kinetic friction i s .........

A block of mass 20 kg is lying on an inclined plane of angle 30^(@). I

1.	A block of mass 20 kg is lying on an inclined plane of angle 30^(@). In order to make it move upward along the slope with an acceleration of 25 cm/s^(2), a horizontal force of 400 N is required to be applied on it. (i) Frictional force on the block is .......N. (ii) Co-efficient of kinetic friction i s .........
Answer» 134,0.56 207,1.52 243.41, 0.66 400,0.42 Solution :Accordingro thefiguretwocomponentsofweightmgaremg cos `theta`andmg sin `theta `and twocomponentsof weight`vec(F )` areF cos `theta`and F sin `theta`Supposeblockmovesupwardson theslopewithacceleration`a_(x )` But`SIGMA F_(x ) = ma_(x )` `F cos30^(@) F -mg sin 30^(@)= ma_(x ) ( :., theta= 30^(2))` `:.346 .4 - f - 98=5` `:. 346 .4 -98-5 = f` Nowf= `mu_(s ) mg` `mu_(s ) = (f )/( mg )` `=(243 .4 )/( 20 XX 9.8) ` `=1.2418 ` `1.24`