Newton's second law: A 50.0-N box is sliding on a rough horizontal floor, and the only horizontal force acting on it is friction. You observe that at one instant the box is sliding to the right at 1.75 m/s and that it stops in 2.25 s with uniform acceleration. What magnitude force does friction exert on this box?
A) 3.97 N
B) 8.93 N
C) 38.9 N
D) 50.0 N
E) 490 N

Question

Quizplus · Accepted Answer

The frictional force can be calculated using the formula: friction force = mass * acceleration. First, find the acceleration using the formula: final velocity = initial velocity + (acceleration * time). Since the box stops, final velocity = 0, initial velocity = 1.75 m/s, and time = 2.25 s. Solving for acceleration gives -0.778 m/s². The mass of the box is weight/gravity = 50.0 N / 9.8 m/s² = 5.10 kg. Therefore, friction force = 5.10 kg * 0.778 m/s² = 3.97 N.

Newton's Second Law: a 50