A force of 36.1 N is required to start a 4.0 -kg box moving across a horizontal concrete floor. If the 36.1 N force continues, the box accelerates at 0.71 m/s^2. What is the coefficient of kinetic friction?
I got 0.072 as my answer and this is showing as in correct.
I did:
36.1 - mu x 4.0kg x 9.8 = 4.0kg x 0.71
0.71 / 9.8 = 0.072
To find the coefficient of kinetic friction, we need to use the given information about the force and acceleration.
First, let's calculate the net force acting on the box. The net force is the difference between the applied force and the force of friction.
The applied force is given as 36.1 N.
The mass of the box is given as 4.0 kg.
The acceleration is given as 0.71 m/s^2.
Using Newton's second law of motion, which states that the net force is equal to the mass multiplied by the acceleration (F = m*a), we can calculate the net force.
Net force = mass * acceleration
Net force = 4.0 kg * 0.71 m/s^2
Net force = 2.84 N
Since the net force is the difference between the applied force and the force of friction, we can set up the following equation:
Net force = Applied force - Force of friction
Substituting the values we know, we have:
2.84 N = 36.1 N - Force of friction
Solving for the force of friction:
Force of friction = 36.1 N - 2.84 N
Force of friction = 33.26 N
Now, we can use the equation for the frictional force:
Force of friction = coefficient of friction * normal force
The normal force is the force exerted by the surface on the box, which is equal to the weight of the box when it is on a horizontal surface.
The weight of the box is given by the equation:
Weight = mass * gravity
where gravity is approximately 9.8 m/s^2.
Weight = 4.0 kg * 9.8 m/s^2
Weight = 39.2 N
Substituting the known values:
33.26 N = coefficient of friction * 39.2 N
Solving for the coefficient of friction:
coefficient of friction = 33.26 N / 39.2 N
coefficient of friction ≈ 0.85
Therefore, the coefficient of kinetic friction is approximately 0.85.
netforce=mass*acceleration
36.1-mu*mass*g=mass*acceleration
solve for mu.