Question

I will Follow the Correct First Answer ______&&&


A 2.5 kg block is initially at rest on a horizontal surface. A 6 N horizontal force and a vertical force P are applied to the block . The Coefficient of static friction for the block and surface is 4 . The magnitude of the friction force when P = 9 N : [ g = 10$\tt{m/s^2}$ ] ​

Answer 1

maximum static force that can be applied on block..is =4×2.5×10

=100N +9=109N (9N of force is applied externally in vertically below direction)

horizontal force=6N

the maximum force ,upto which there will not be any sliding is 109N,since it is the maximum static friction......

it shows on applying,a force of 6N , doesn't create any effect on block and it will be remain at the same position...

frictional force will be same of 6N in opposite direction ,of the force applied......

be brainly ✌️

keep asking ur doubts.....

Answer 2

Answer:

hey there here is ur answer....

u can solve it by finding out the limiting force of friction and then compare with the given force...

if u need more hint then u can prefer examples of hcv