A 3.0 kg block is being pulled with a horizontal force of 29.7 N. If the block is accelerating at 3.2 m/s/s, what is the coefficient of kinetic friction between the block and the table?

To find the coefficient of kinetic friction between the block and the table, we can use Newton's second law of motion and the formula for friction.

The equation for Newton's second law of motion is Fnet = ma, where Fnet is the net force acting on an object, m is the mass of the object, and a is the acceleration of the object.

In this case, the net force acting on the block is the force applied to pull the block minus the force of kinetic friction. So we can write it as:

Fnet = Fapplied - Ffriction

Where Ffriction is the force of kinetic friction.

The formula for the force of kinetic friction is:

Ffriction = μk * N

Where μk is the coefficient of kinetic friction and N is the normal force.

The normal force is the force exerted by the surface (in this case, the table) perpendicular to the object. In this case, the normal force is equal to the weight of the block, which is given by:

N = mg

Where m is the mass of the block and g is the acceleration due to gravity (approximately 9.8 m/s^2).

Substituting the values into the equations, we have:

Fnet = ma

Fapplied - Ffriction = ma

Fapplied - μk * N = ma

Fapplied - μk * mg = ma

We are given the mass of the block (m = 3.0 kg), the applied force (Fapplied = 29.7 N), and the acceleration (a = 3.2 m/s^2). We can substitute these values into the equation to solve for the coefficient of kinetic friction (μk).

Let's solve the equation:

29.7 N - μk * (3.0 kg)(9.8 m/s^2) = (3.0 kg)(3.2 m/s^2)

29.7 N - 29.4 μk = 9.6 N

Simplifying the equation:

-29.4 μk = 9.6 N - 29.7 N

-29.4 μk = -20.1 N

Dividing both sides of the equation by -29.4:

μk ≈ -20.1 N / -29.4 N ≈ 0.683

Therefore, the approximate coefficient of kinetic friction between the block and the table is 0.683.