A force of 23.0 N is required to start a 3.1 kg box moving across a horizontal concrete floor.
(a) What is the coefficient of static friction between the box and the floor?
(b) If the 23.0 N force continues, the box accelerates at 0.50 m/s2. What is the coefficient of kinetic friction?
How do I find the answer to part b?
applied force-friction=mass*acceleration
solve for friction, then set it equal to mu*mg and solve for mu.
so i would do.. 23N - friction= 3.1(or 30.38N) *.5
and get f=7.81
then 7.81 = mu k * mu g ?
To find the coefficient of kinetic friction in part (b), you can use the equation that relates the force of kinetic friction (Fk), the mass of the box (m), and the acceleration of the box (a):
Fk = μk * m * g
Where:
- Fk is the force of kinetic friction.
- μk is the coefficient of kinetic friction.
- m is the mass of the box.
- g is the acceleration due to gravity (approximately 9.8 m/s²).
In this case, we need to rearrange the equation to solve for μk:
μk = Fk / (m * g)
Given information:
- Force of kinetic friction (Fk) = 23.0 N (same as in part a).
- Mass of the box (m) = 3.1 kg (same as in part a).
- Acceleration of the box (a) = 0.50 m/s².
Since the box is accelerating, the applied force is greater than the force of kinetic friction. By using Newton's second law, we can find the force of kinetic friction:
Fk = m * a
Plugging in the values:
Fk = 3.1 kg * 0.50 m/s²
Fk = 1.55 N
Now, we can substitute the known values into the equation for the coefficient of kinetic friction:
μk = Fk / (m * g)
μk = 1.55 N / (3.1 kg * 9.8 m/s²)
μk ≈ 0.051
Therefore, the coefficient of kinetic friction, rounded to three decimal places, is approximately 0.051.