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 .........

Options

A. 134,0.56
B. 207,1.52
C. 243.41, 0.66
D. 400,0.42

Answer

C

Explanation

According ro the figure two components of weight mg are mg cos `theta` and mg sin `theta ` and two components of weight `vec(F ) ` are F cos `theta` and F sin `theta` Suppose block moves upwards on the slope with acceleration `a_(x )`
But `Sigma F_(x ) = ma_(x )`
`F cos 30^(@) F - mg sin 30^(@) = ma_(x ) ( :. , theta = 30^(2))`
`:. 346 .4 - f - 98 =5`
`:. 346 .4 -98 -5 = f`
Now f= `mu_(s ) mg`
`mu_(s ) = (f )/( mg )`
`=(243 .4 )/( 20 xx 9.8) `
`=1.2418 `
`1.24`