HELP ME WITH PHYSICS!!!

Question

A mop is pushed across the floor with a force 𝐹 of 40.8 N at an angle of 𝜃=51.2∘. The mass of the mop head is 𝑚=2.10 kg. Calculate the magnitude of the acceleration 𝑎of the mop head if the coefficient of kinetic friction between the mop head and the floor is 𝜇𝑘=0.360.

Answer 1

Given:

𝐹 = 40.8 N at an angle of 𝜃=51.2

Mass m = 2.10kg

µ_k= 0.360

Find: Calculate the magnitude of the acceleration 𝑎 of the mop head

The mop head is moving horizontally. It is not moving vertically. There is no vertical acceleration.

To calculate the acceleration of the mop head. You need to find the sum of the forces in the horizontal direction.

Let F_p = the push force

Let F_k = the force due to kinetic friction.

Let the positive direction be to the right.

When the mop is pushed to the right, the kinetic friction force is opposing it and pushing to the left.

The mop is being pushed at an angle. When we determine the sum of the forces in the horizontal direction, we only use the F_{px =} 40.8 cos (51.2)

The formula for F_k= µ_kF_n

F_N is the normal force. In this case, since the mop is being pushed at an angle, it will be the force in the y direction due to gravity plus the y component of the force pushing on the mop

F_N= mg + 40.8sin(51.2)N

F_N= (2.10kg)(9.81m/s²) + (40.8N)sin(51.2)

F_k= µ_kF_N= (0.360)[(2.10kg)(9.81m/s²) + (40.8N)sin(51.2)]

F=ma

The sum of the forces in the horizontal direction would be

F_px - F_k = ma_x

40.8N cos (51.2) - (0.360)[(2.10kg)(9.81m/s²) + (40.8N)sin(51.2)] = (2.10kg)(a_x)

Now you can solve for a_x

a_x=3.19