A force of 30.0 N is required to start a 3.8 kg box moving across a horizontal concrete floor

If the 30.0 N force continues, the box accelerates at 0.50 m/s2. What is the coefficient of kinetic friction?

- (0.5m/s^2)*(3.8kg) = 1.9N

Because Force = (mass)*(acceleration)
- Force of Friction = (30N)-(1.9N) = 28.1N
Because the 30N force is being accelerated therefore the force is does not have to be as strong so the force ultimately decreases.
- Force of Surface = (3.8kg)(9.8m/s^2) = 37.24N
28.1N = (37.24N)*(mu)
mu = 0.755
Because Force of Friction = (Coefficient of Friction)*(Force of Surface)

To find the coefficient of kinetic friction, we can use the formula:

Coefficient of kinetic friction (μk) = (Net force due to friction) / (Normal force)

First, let's find the net force due to friction. The net force (Fnet) is equal to the mass (m) multiplied by the acceleration (a):

Fnet = m * a

Given:
Force (F) = 30.0 N
Mass (m) = 3.8 kg
Acceleration (a) = 0.50 m/s^2

Fnet = 3.8 kg * 0.50 m/s^2
= 1.9 N

Next, we need to find the normal force (Fn) acting on the box. The normal force is the force exerted by a surface to support the weight of an object resting on it. In this case, the normal force is equal to the weight of the box since it is resting on a horizontal surface.

Weight (W) = mass * gravitational acceleration
= m * g

Given:
Mass (m) = 3.8 kg
Gravitational Acceleration (g) = 9.8 m/s^2

W = 3.8 kg * 9.8 m/s^2
= 37.24 N

Since the vertical load is equal to the norma force, Fn = 37.24 N.

Finally, we can calculate the coefficient of kinetic friction:

Coefficient of kinetic friction (μk) = Fnet / Fn
= 1.9 N / 37.24 N

Calculating the value:

μk = 0.051

Therefore, the coefficient of kinetic friction is approximately 0.051.

To find the coefficient of kinetic friction, we need to use Newton's second law of motion and the equation for the force of kinetic friction:

F_net = m * a

First, let's calculate the net force acting on the box. The net force is the force that accelerates the box:

F_net = F_applied - F_friction

Where:
F_applied = the applied force on the box (which is 30.0 N)
F_friction = the force of kinetic friction

Now, we can rewrite Newton's second law as:

F_net = m * a

Substituting the values:

F_applied - F_friction = m * a

Rearranging the equation to solve for F_friction:

F_friction = F_applied - m * a

Plugging in the given values:

F_friction = 30.0 N - (3.8 kg * 0.50 m/s^2)

Calculating the frictional force:

F_friction = 30.0 N - 1.9 N

F_friction = 28.1 N

The force of kinetic friction is 28.1 N.

Next, we can calculate the coefficient of kinetic friction using the equation:

F_friction = μ_k * N

Where:
μ_k = coefficient of kinetic friction
N = Normal force

The normal force is the force exerted by the surface on the box vertically upward. Since the box is on a horizontal surface without any vertical acceleration, the normal force is equal to the weight of the box.

N = m * g

Where:
m = mass of the box (3.8 kg)
g = acceleration due to gravity (9.8 m/s^2)

Substituting the values:

N = 3.8 kg * 9.8 m/s^2

Calculating the normal force:

N = 37.24 N

Now, we can substitute the values of F_friction and N into the equation for coefficient of kinetic friction:

28.1 N = μ_k * 37.24 N

Solving for μ_k:

μ_k = 28.1 N / 37.24 N

Calculating the coefficient of kinetic friction:

μ_k ≈ 0.754

Therefore, the coefficient of kinetic friction is approximately 0.754.

First of all, the static coefficient of friction is

mus = 30/(3.8*9.8)= 0.806

After motion starts, the friction force f is given by:
(30 - f) = M a = 9.8 * 0.5 = 4.9 N
f = 25.1 N
muk = f/(M*g) = 25.1/(9.8*3.8) = 0.675