A physics student is pulling on a 20 kg box that is sliding on the floor. The student applies a pulling force of 100 N using a rope at angle 30o
from the horizontal. The coefficient of kinetic friction between the box and the floor is 0.4. The acceleration of the box is closest to?
Thank you Daemon for all your help! :)
100 cos 30 = horizontal pull = 86.6 N
100 sin 30 = vertical pull = 50 N
Normal force = mg - 50 = 146 N
Friction Force = .4*146 = 58.5 N
total horizontal force = 86.6-58.5 = 28.1 N
F = m a
28.1 = 20 a
a = 1.41 m/s^2
To find the acceleration of the box, we need to consider the forces acting on it. There are two main forces: the pulling force applied by the student and the kinetic friction force opposing the motion.
1. Calculate the horizontal component of the pulling force:
F_horizontal = F_pull * cos(angle)
F_horizontal = 100 N * cos(30°)
2. Calculate the friction force:
F_friction = coefficient_friction * normal_force
The normal force is equal to the weight of the box, which is given by:
normal_force = mass * gravitational_acceleration
normal_force = 20 kg * 9.8 m/s^2
F_friction = 0.4 * (20 kg * 9.8 m/s^2)
3. Calculate the net force in the horizontal direction:
net_force = F_horizontal - F_friction
4. Calculate the acceleration of the box using Newton's second law:
net_force = mass * acceleration
acceleration = net_force / mass
Substituting the values, we can find the acceleration of the box.