When you apply a horizontal force of 76 N to a block, the block moves across the floor at a constant speed ().When you apply a force of 89 N, what is the magnitude of the horizontal component of the force that the floor exerts on the block?

Explanation:

When we apply a horizontal force of 76 N to a block, the block moves across the floor at a constant speed. So net force on the block is zero .

It implies that a force ( frictional ) acts on it which is equal to 76 N in opposite direction ( friction )

When we apply  a greater force on it it starts moving with acceleration .

This time kinetic friction acts on it due to rough ground equal to 76 N .This is limiting friction ( maximum friction )

Net force on the body in later case

= 89 - 76

= 13 N

Force by ground on the block in horizontal direction = 76 N ( FRICTIONAL FORCE )

=

trig 15.3 cos

a and c

c is 100% but i am confused in a