Question

A 26.0 kg box is being pushed by a person across a horizontal floor with a force F = 135 N that makes an angle θ = 27.0° with the horizontal as shown in the figure. What is the magnitude of the acceleration of the box if the coefficient of kinetic friction between the box and the floor is 0.15?

Answer 1

To find the magnitude of the acceleration of the box, we need to consider the forces acting on it. The applied force, F, is the horizontal component of the force applied by the person, which is given by F = 135 N * cos(27°). The weight of the box, mg, is equal to 26.0 kg * 9.8 m/s².

Step-by-step explanation:

To find the magnitude of the acceleration of the box, we need to consider the forces acting on it. The applied force, F, is the horizontal component of the force applied by the person, which is given by F = 135 N * cos(27°). The weight of the box, mg, is equal to 26.0 kg * 9.8 m/s². The frictional force, ff, can be found using the coefficient of kinetic friction, which is 0.15. The equation for friction where N is the normal force.

Using these values, we can now calculate the net force on the box:

• Net force (Fx) = F - ff
• Net force (Fy) = mg - N
• Acceleration (ax) = Net force (Fx) / mass

Substituting the values, we get:

• Net force (Fx) = (135 N * cos(27°)) - (0.15 * 26.0 kg * 9.8 m/s²)
• Net force (Fy) = (26.0 kg * 9.8 m/s²) - N
• Acceleration (ax) = (Net force (Fx)) / 26.0 kg

Calculating these values will give us the magnitude of the acceleration of the box.