a force of 400 N Pushes on a 30 kg box. Starting from rest, the box attains a speed of 4 m/s in 5 seconds. determine the mu between the box and floor.
net force= ma
400-forcefriction=ma
400-mu*mg=ma
where a= 4/5 m/s^2
solve for mu
I've tried over and over and get mu = .36 which I know is wrong. (Supposed to be .51)
What am I doing wrong?
To determine the coefficient of friction (μ) between the box and the floor, we need to use Newton's second law of motion and the equation for the force of friction.
First, let's calculate the acceleration of the box using Newton's second law:
Force = mass × acceleration
Rearranging the formula to solve for acceleration:
Acceleration = Force / mass
Acceleration = 400 N / 30 kg
Acceleration = 13.33 m/s²
Next, we can use the equation of motion to find the distance traveled by the box in 5 seconds:
Distance = (initial velocity × time) + (0.5 × acceleration × time²)
Since the box starts from rest (initial velocity = 0), the equation simplifies to:
Distance = 0.5 × acceleration × time²
Distance = 0.5 × 13.33 m/s² × (5 s)²
Distance = 0.5 × 13.33 m/s² × 25 s²
Distance = 166.625 m
Now, let's calculate the force of friction acting on the box. At constant velocity, the force of friction is equal in magnitude but opposite in direction to the applied force:
Force of friction = applied force
Force of friction = 400 N
Lastly, we can calculate the normal force (N) using the equation:
Normal force = mass × gravity
Normal force = 30 kg × 9.8 m/s²
Normal force = 294 N
The force of friction can be written as:
Force of friction = μ × N
Since the box is not moving vertically, the normal force is equal to the weight of the box. Thus, we can substitute N with 294 N:
400 N = μ × 294 N
Now we can solve for μ:
μ = 400 N / 294 N
μ ≈ 1.36
Therefore, the coefficient of friction (μ) between the box and the floor is approximately 1.36.