Question

(5)
A block of mass 15 kg is lying on an inclined plane
of angle 30°. In order to make it more 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 is
(g = 9.8 ms=2)
(A) 99.70 N (B) 134 N (C) 90 N (D) 95.95 N​

Answer 1

Given info : A block of mass 15 kg is lying on an inclined plane of angle 30°. in order to make it more upward along the slope with an acceleration 25 cm/s², a horizontal force of 200N is required to be applied on it.

To find : the frictional force on the block is ...

solution : force along plane = Fcos30° (upward)

weight of block along plane = mgsin30° (downward)

block goes upward so frictional force acts downward direction.

from Newton's law,

upward force - downward force = mass × acceleration

⇒Fcos30° - fr - mgsin30° = ma

⇒200N × √3/2 - fr - 15 kg × 9.8 m/s² × 1/2 = 15kg × 0.25 m/s²

⇒173.2 N - fr - 73.5 N = 3.75 N

⇒fr = 95.95 N

Therefore frictional force applied on it is 95.95 N

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Answer 2

i hope the above given answer will help you...