A physics student is pulling on a 20 kg box that is sliding on the floor. The student applies a pulling force of 100 N using a rope at angle 30o

from the horizontal. The coefficient of kinetic friction between the box and the floor is 0.4. The acceleration of the box is closest to?

ma=Fcosα-F(fr)

F(fr) =μmg
a= (Fcosα- μmg)/m

To determine the acceleration of the box, we need to analyze the forces acting on it.

1. First, let's find the normal force (N) exerted on the box by the floor. The normal force is equal in magnitude and opposite in direction to the weight of the box. The weight (W) can be calculated as the mass (m) multiplied by the acceleration due to gravity (g), which is approximately 9.8 m/s^2.
W = m * g
W = 20 kg * 9.8 m/s^2
W = 196 N

Therefore, the normal force is 196 N.

2. Next, let's determine the force of friction (F_friction) acting on the box. The force of friction can be calculated using the coefficient of kinetic friction (μ_k) and the normal force (N).
F_friction = μ_k * N
F_friction = 0.4 * 196 N
F_friction = 78.4 N

Therefore, the force of friction is 78.4 N.

3. Now, let's decompose the applied pulling force into horizontal and vertical components. Since the angle of the rope is given as 30 degrees from the horizontal, the horizontal component (F_horizontal) can be found using trigonometry.
F_horizontal = F_applied * cos(angle)
F_horizontal = 100 N * cos(30 degrees)
F_horizontal ≈ 86.6 N

Therefore, the horizontal component of the applied force is approximately 86.6 N.

4. The net force acting on the box in the horizontal direction can be calculated by subtracting the force of friction from the horizontal component of the applied force.
Net force (F_net) = F_horizontal - F_friction
F_net = 86.6 N - 78.4 N
F_net ≈ 8.2 N

Therefore, the net force acting on the box in the horizontal direction is approximately 8.2 N.

5. Finally, we can use Newton's second law of motion (F = m * a) to find the acceleration (a) of the box. Rearranging the equation, we have:
a = F_net / m
a = 8.2 N / 20 kg
a ≈ 0.41 m/s^2

Therefore, the acceleration of the box is approximately 0.41 m/s^2.