A 50-kg box is being pushed along a horizontal surface. The coefficient of static friction between the box and the ground is 0.65, and the coefficient of kinetic friction is 0.35. What horizontal force
Must be exerted on the box for it to accelerate at 1.2 m/s²?
A) 120 N
B) 170 N
C) 60 N
D) 490 N
E) 230 N

Question

Quizplus · Accepted Answer

To find the force required to accelerate the box, we first calculate the force of static friction since it needs to be overcome to start moving the box. The force of static friction (F_static) is given by F_static = μ_static * N, where μ_static is the coefficient of static friction and N is the normal force. The normal force (N) is equal to the weight of the box, which is mass (m) times gravity (g), so N = m * g. For a 50-kg box, N = 50 kg * 9.8 m/s² = 490 N. Therefore, F_static = 0.65 * 490 N = 318.5 N.Once the box starts moving, we deal with kinetic friction to maintain acceleration. The force of kinetic friction (F_kinetic) is given by F_kinetic = μ_kinetic * N, where μ_kinetic is the coefficient of kinetic friction. So, F_kinetic = 0.35 * 490 N = 171.5 N.The net force (F_net) required to accelerate the box is the sum of the force to overcome kinetic friction and the force needed to accelerate the box at 1.2 m/s², given by Newton's second law, F = m * a, where a is the acceleration. So, the additional force needed for acceleration is F_acceleration = 50 kg * 1.2 m/s² = 60 N.Therefore, the total force required is F_total = F_kinetic + F_acceleration = 171.5 N + 60 N = 231.5 N, which rounds to 230 N, making E the correct answer.

A 50-Kg Box Is Being Pushed Along a Horizontal Surface