In the given arrangement, the normal force applied by block on the ground is

The normal force applied by block on the ground is (mg - F cosФ)

Explanation:

Lets explain how to solve the problem

At first you must distribute the force F into two components

Vertical component which is F cosФ

Horizontal component which is F sinФ

The block is in equilibrium, that means sum of forces acting on the

block is zero

So the upward forces equal the downward forces

Normal reaction force R applied by block on the ground and the

vertical component of F both are upward forces

The weight of the block is downward force

The normal reaction force R plus the vertical component of F is

equal to the weight

R + F cosФ = W

W = mg, where g is acceleration of gravity and m is the mass of

the block

R + F cosФ = mg

Subtract F cosФ from both sides

R = mg - F cosФ

The normal force applied by block on the ground is (mg - F cosФ)

i think the answer is a, hope this is right, that would be my best guess.

be subject to a force andchange its motion