One day while moving boxes you get tired and decide to use a rocket instead. You attach a small rocket of negligible mass to a 6.45 kg box. When you turn the rocket on, it provides a constant thrust of 212 N, and the sbox begins sliding across the pavement. If the magnitude of acceleration of the box is 17.0 m/s2, what is the coefficient of kinetic friction between the soapbox and pavement?

To find the coefficient of kinetic friction between the soapbox and the pavement, we can use Newton's second law of motion, which states that the net force acting on an object is equal to the product of its mass and acceleration.

First, let's find the net force acting on the box. The only forces acting on the box are the thrust force provided by the rocket and the force of kinetic friction opposing its motion.

Net force (F_net) = Thrust force (F_thrust) - Force of friction (F_friction)

Since the mass of the box is given as 6.45 kg and the acceleration is given as 17.0 m/s^2, we can calculate the net force as follows:

F_net = m * a
F_net = 6.45 kg * 17.0 m/s^2
F_net = 109.65 N

Now we need to find the force of friction. The force of friction is calculated using the equation:

F_friction = coefficient of kinetic friction (μ_k) * Normal force (N)

Since the box is sliding on the horizontal pavement, the normal force is equal to the weight of the box, which is given by:

N = m * g
N = 6.45 kg * 9.8 m/s^2
N = 63.21 N

Substituting the given thrust force and net force into the equation, we get:

109.65 N = 212 N - μ_k * 63.21 N

Now, rearranging the equation to solve for the coefficient of kinetic friction:

μ_k = (212 N - 109.65 N) / 63.21 N

Calculating the value:

μ_k = 102.35 N / 63.21 N
μ_k ≈ 1.618

Therefore, the coefficient of kinetic friction between the soapbox and the pavement is approximately 1.618.