A stockroom worker pushes a box with 12.3 kg on a horizontal surface of 3.10 m/s . The coef­ficients of kinetic and static friction between the box and the surface are 0.280 and 0.480, respectively.

Question

Part A

What horizontal force must the worker apply to maintain the motion of the box?

Part B

If the worker stops pushing, what will be the acceleration of the box?

Answer 1

Part A.

There are 4 forces acting on the body:

1. Weight mg, whose direction is vertically down
2. Normal reaction force N perpendicular to the surface
3. Force F exerted by the worker in the direction of motion and parallel to the surface
4. Friction f in the direction opposite to the motion of the body

Let's select the coordinate so that x -axis is in the direction of motion and y-axis perpendicular to it. Then the Second Newton's law in scalar form (in case of constant speed there is no acceleration) looks like

x: F - f = 0 (1)

y: N - mg = 0 (2)

Knowing that frictional force f = µ_kN and using (2), we will get f =µ_kmg . Because the body is moving, we need to take the coefficient of kinetic friction μ_k.

Then we can rewrite equation (1) as

F - µ_kmg = 0

Expressing for force, that the worker exerts, we will have

F = µ_kmg

Plugging in the values of all variables, we obtain

F = 0.280 x 12.3 kg x 9.8 m/s² = 33.8 N

Answer: 33.8 N

Part B.

If the worker stops pushing, then the only unbalanced force that causes deceleration of the body is kinetic friction force, and we can write:

f = ma

Hence, acceleration a = f/m or a = F/m, because the magnitude of friction is the same as external force from the worker in case A.

Putting the numbers, we get:

a = 33.8 N / 12.3 kg = 2,7 m/s²

Keep in mind that the direction of the acceleration is opposite to the direction of motion of the box.

Answer: 2,7 m/s² in the direction opposite to the direction of motion of the box