a 60 kilogram student sitting on a hardwood floor accelerates at 1.3 meters per second squared when pulled by a 240 newton horizontal force. what is the coefficient of friction between the student and the floor?
The net accelerating force is F = M a = 78 N. That means the friction force is
Ff = 240 - 78 = 162 N
The coefficient of friction is
(162 N)/(M*g) = ___
To find the coefficient of friction between the student and the floor, we can start by calculating the net force acting on the student.
The net force can be found using Newton's second law of motion, which states that the net force is equal to the product of mass (m) and acceleration (a):
net force = mass × acceleration
Given:
Mass of the student (m) = 60 kg
Acceleration (a) = 1.3 m/s²
Substituting the values into the formula, we have:
net force = 60 kg × 1.3 m/s²
net force = 78 N
Next, we need to find the force of friction (Ff) acting on the student. The force of friction can be calculated using the equation:
force of friction (Ff) = coefficient of friction (μ) × normal force (Fn)
The normal force (Fn) is equal to the weight of the student, which can be calculated as:
Fn = mass × gravity
Fn = 60 kg × 9.8 m/s²
Fn = 588 N
Substituting the values into the equation for the force of friction:
78 N = μ × 588 N
Now, we can solve for the coefficient of friction (μ):
μ = 78 N / 588 N
Calculating the division, we find:
μ ≈ 0.133
Therefore, the coefficient of friction between the student and the floor is approximately 0.133.