A 50.0-kg box is resting on a horizontal floor. A force of 250 N directed at an angle of 20.7° below the horizontal is applied to the box. The coefficient of kinetic friction between the box and the surface is 0.300. What is the acceleration of the box? A) 1.21 m/s² B) 1.77 m/s² C) 2.84 m/s² D) 3.54 m/s² E) 5.14 m/s²

Question

Quizplus · Accepted Answer

The horizontal component of the applied force is 250 N * cos(20.7°). The normal force is increased by the vertical component of the applied force, 250 N * sin(20.7°), and the weight of the box. The frictional force is the coefficient of kinetic friction times the normal force. Subtract the frictional force from the horizontal component of the applied force to find the net force, then use F = ma to find the acceleration.

A 500-Kg Box Is Resting on a Horizontal Floor