(b) 1.77 m/s² (although the actual value is slightly higher) is the correct option.
To solve this problem, we need to consider the forces acting on the box and analyze their components:
Applied force (F): 250 N directed 30° below the horizontal. We can break this force into horizontal and vertical components:
Horizontal component (Fh): Fh = F * cos(30°) = 250 N * cos(30°) ≈ 216.5 N (acts in the direction of motion).
Vertical component (Fv): Fv = F * sin(30°) = 250 N * sin(30°) ≈ 125 N (acts downwards but won't affect horizontal motion as the surface is horizontal).
Weight of the box (W): 50 kg * 9.81 m/s² ≈ 490.5 N (acting downwards).
Friction force (Ff): Opposes the motion, proportional to the normal force (Fn) and the coefficient of kinetic friction (μk): Ff = μk * Fn.
Steps to solve:
Normal force: Since the vertical component of the applied force is balanced by the normal force from the surface, Fn = Fv ≈ 125 N.
Friction force: Ff = μk * Fn = 0.30 * 125 N ≈ 37.5 N (acting in the opposite direction of Fh).
Net force: The net force accelerating the box is the horizontal component of the applied force minus the friction force: Fnet = Fh - Ff ≈ 216.5 N - 37.5 N ≈ 179 N.
Acceleration: Finally, we can calculate the acceleration (a) using Newton's second law: a = Fnet / m = 179 N / 50 kg ≈ 3.58 m/s².
Therefore, the closest answer choice to the acceleration of the box is: