A 10kg box is pulled a distance of 100m along a level floor at constant velocity using a force of 40 N at an angle of 35 degrees above the horizontal. If the box is pulled at constant velocity, what is the coefficient of friction between the bottom of the box and the floor?
Force horizontal=40cos35
That will equal friction
Force normal=mg-40cos35
friction=forcehorizontal
mu(mg-40sin35)=40Cos35
solvefor mu
To find the coefficient of friction between the bottom of the box and the floor, we can use Newton's second law of motion.
First, let's resolve the force into horizontal and vertical components. The vertical component of the force can be found using the formula: F_vertical = F * sin(angle). Substituting the given values, we have F_vertical = 40 N * sin(35 degrees) = 23 N.
Since the box is moving at constant velocity, the net force acting on it must be zero. This means that the force of friction must be equal in magnitude and opposite in direction to the horizontal component of the applied force.
The horizontal component of the force can be found using the formula: F_horizontal = F * cos(angle). Substituting the given values, we have F_horizontal = 40 N * cos(35 degrees) = 32.71 N.
Now, let's calculate the force of friction. We have F_friction = F_horizontal = 32.71 N.
The force of friction can be calculated using the formula: F_friction = coefficient of friction * normal force. The normal force is equal to the weight of the box, which can be calculated using the formula: weight = mass * gravity. Substituting the given values, we have weight = 10 kg * 9.8 m/s^2 = 98 N.
Now, substituting the values into the equation F_friction = coefficient of friction * normal force, we have 32.71 N = coefficient of friction * 98 N.
Dividing both sides of the equation by 98 N, we have coefficient of friction = 32.71 N / 98 N.
Simplifying, the coefficient of friction between the bottom of the box and the floor is approximately 0.334.
Therefore, the coefficient of friction between the bottom of the box and the floor is approximately 0.334.