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)
