A block of mass 5 kg is moving on an inclined plane which makes an angle of 30 with the horizontal. Friction coefficient between the block and inclined plane surface is 32. The force to be applied on the block so that the block will move down without acceleration is N. (g = 10 m/s2).
- 25
- 12.5
- 15
- 7.5
Answer
(2) 12.5
Solution
Component of weight along the incline: mg 30 = 5 10 12 = 25 N (down the incline)
Kinetic friction (up the incline, block moving down): mg 30 = 32 50 32 = 37.5 N
Friction exceeds mg 30 , so a force F must act down the incline.
No acceleration F + 25 = 37.5
F = 12.5 N
Answer: (2)