A 50.0-kg box is being pushed along a horizontal surface by a force of 250 n directed 28.0o below the horizontal. the coefficient of kinetic friction between the box and the surface is 0.300. what is the acceleration of the box?

Refer to the figure shown below.

Tha applied force of F = 250 N has

(i) a horizontal component of Fx = (250 N) cos (28°) = 220.737 N,

(ii) an upward vertical component of Fy = (250 N) sin (28°) = 117.368 N

The weight of the box is

W = (50 kg) * (9.8 m/s²) = 490 N

The normal reaction on the box is

N = W - Fx = 490 - 117.368 = 372.632 N

The resistive frictional force on the box is

μN = 0.3*372.632 = 111.790 N

The horizontal driving force on the box is

Fx - μN = 220.737 - 111.790 = 108.947 N

If the acceleration of the box is a m/s², then

(50 kg) * (a m/s²) = (108.947 N)

a = 108.947/50 = 2.179 m/s²

Answer:

The acceleration of the box is 2.18 m/s² (nearest hundredth)