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?
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)
