Question

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?

Answer 1

To find the acceleration of the box, we need to consider the forces acting on it. There is a horizontal force of 250 N pushing the box, and there is a frictional force opposing its motion. The normal force can be calculated using the formula: normal force = mass * gravitational acceleration * cos(theta). Once we have the frictional force and the net force, we can use Newton's second law to find the acceleration: net force = mass * acceleration.

Step-by-step explanation:

To find the acceleration of the box, we need to consider the forces acting on it. There is a horizontal force of 250 N pushing the box, and there is a frictional force opposing its motion. The frictional force can be calculated using the formula:

frictional force = coefficient of kinetic friction * normal force

The normal force can be calculated using the formula:

normal force = mass * gravitational acceleration * cos(theta)

Once we have the frictional force and the net force, we can use Newton's second law to find the acceleration:

net force = mass * acceleration

Plugging in the values, we can calculate the acceleration of the box.

Answer 2

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)