A 75kg box is being pulled across the floor with a force of 465N.If the kinetic coefficient of friction is 0.57,what is the acceleration of the box?
Box weight = m g = 735 N
Friction force = 0.57*735 = 419 N
net force = Fnet = 465 - 419 = 46 N
acceleration = Fnet/m = 46/75
= 0.613 m/s^2
To find the acceleration of the box, we need to use Newton's second law of motion, which states that the force acting on an object is equal to the mass of the object multiplied by its acceleration.
First, let's calculate the force of friction that opposes the motion of the box. The force of friction can be calculated using the formula:
Force of friction = (kinetic coefficient of friction) * (normal force)
The normal force is the perpendicular force exerted by the surface on the object. In this case, since the box is on the floor, the normal force is equal to the weight of the box, which can be calculated using the formula:
Weight = mass * acceleration due to gravity
Acceleration due to gravity is approximately 9.8 m/s^2.
Weight = 75 kg * 9.8 m/s^2 = 735 N
Now, we can calculate the force of friction:
Force of friction = (0.57) * (735 N) = 419.95 N
Since the force of friction opposes the motion, we need to use the force of friction as a negative value. So, the force acting on the box can be calculated as:
Net force = (force applied) - (force of friction)
Net force = 465 N - 419.95 N = 45.05 N
Finally, we can calculate the acceleration using Newton's second law:
Net force = mass * acceleration
acceleration = Net force / mass
acceleration = 45.05 N / 75 kg = 0.6 m/s^2
Therefore, the acceleration of the box is 0.6 m/s^2.