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A force of 250 n pushes a 50-kg box along a horizontal surface. the force is directed 30o below the horizontal. what is the acceleration of the box if the coefficient of kinetic friction between the box and the surface is 0.30?

a. 0.637 m/s²
b. 1.77 m/s²
c. 3.16 m/s²
d. 6.31 m/s²

User Edsandorf
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(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:

User AVarf
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