When emma pushes a bag with a force of 27 newtons, the coefficient of kinetic friction between the bag and the floor is 0.23. what is the normal force acting on the bag by the floor?

Nc = 117.4 N

Explanation:

Given:-

- The applied force, F = 27 N

- The coefficient of kinetic friction, uk = 0.23

Find:-

What is the normal force acting on the bag by the floor?

Solution:-

- Develop a FBD of the bag and mark all the forces acting on the bag. With x direction is parallel to the surface and y coordinate normal to the surface.

- There are four forces acting on the bag.

1 ) The applied force by Emma (F) acting in the right ( positive x - direction ).

2) The weight (W) of the bag vertically downwards ( negative y - direction )

3) The normal contact force (Nc) opposite in direction to weight of bag.

4) The frictional force (Ff)  acting in opposite direction to the Emma push.

- The Normal force exerted by the surface on the bag is related to the frictional force Ff as follows:

                                 Ff = uk*Nc

- We will apply the Newton's second law of motion in the x direction as follows. Assuming that bag moves with constant speed or does not accelerate then we have:

                                 F - Ff = m*a

Where, a = 0,

                                 F - Ff = 0

                                 Ff = F = 27 N.

- Now use the Normal Force and Frictional Force relationship as follows:

                                 Nc = Ff / uk

                                 Nc = 27 / 0.23

                                 Nc = 117.4 N

the answer is b ) square root of 5

not moving. pos accel,