Question

A force F=i+ 4j acts on the block shown the force of friction acting on the block.Please explain

Answer 1

Force Equs for 2 blocks are:
at - f = mA (1)
f = mA (2)
A is the acc. of 2 blocks before relative sliding.Solving above eqs. we get
f = at/2
Now when relative sliding starts then
f = Nk = mgk
at = mgk
or t = 2mgk/a

Answer 2

Surely the block will not move and thus the force of friction will be -icap.

.

Explanation.

FORCES IN Y - AXIS...

Force on the block in downward direction is 10N

And acc to question .

In upward direction is 4N

Therefore the net force is in downward direction and it is 6N .so the block will not fly

.

FORCES IN X-AXIS

Given force is 1N.

And maximum frictional force is 3N(by umg)

Therefore the block will not move.

We know that static friction is self adjusting i.e opp to the applied force if the block doednt move.