You push a 1.3 kg block against a horizontal spring, compressing the spring by 22 cm. Then you release the block, and the spring sends it sliding across a tabletop. It stops 79 cm from where you released it. The spring constant is 220 N/m. 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 need to use the information provided in the problem.

First, let's analyze the forces acting on the block when it is released from the compressed spring. At the moment of release, the only horizontal force acting on the block is the force exerted by the spring, which is given by Hooke's Law:

F_spring = -k * x

Where:
F_spring is the force exerted by the spring
k is the spring constant
x is the compression or extension of the spring from its equilibrium position

In this problem, the spring constant (k) is given as 220 N/m, and the compression of the spring (x) is 22 cm or 0.22 m. Therefore, we can calculate the force exerted by the spring:

F_spring = -220 N/m * 0.22 m = -48.4 N

Here, the negative sign indicates that the spring force is acting in the opposite direction to the motion.

The force due to kinetic friction (F_friction) acting in the opposite direction of the motion can be calculated using Newton's second law:

F_friction = m * a

Where:
m is the mass of the block
a is the acceleration of the block

The acceleration of the block can be obtained using the equation of motion for constant acceleration:

v^2 = u^2 + 2 * a * s

Where:
v is the final velocity of the block
u is the initial velocity of the block (which is zero in this case)
s is the displacement of the block

Here, the final velocity (v) is zero because the block comes to a stop, and the displacement (s) is given as 79 cm or 0.79 m. Rearranging the equation, we can solve for the acceleration:

a = -u^2 / (2 * s) = -0 / (2 * 0.79) = 0 m/s^2

Since the acceleration is 0 m/s^2, this tells us that the forces acting on the block are balanced, and there is no net force when it comes to a stop. Therefore, the force due to kinetic friction (F_friction) is equal in magnitude but opposite in direction to the spring force (F_spring):

F_friction = -F_spring = 48.4 N

Now, we can calculate the normal force (N) exerted by the table on the block. The normal force is equal in magnitude and opposite in direction to the weight of the block:

N = mg

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

The mass of the block is given as 1.3 kg, so we can calculate the normal force:

N = 1.3 kg * 9.8 m/s^2 = 12.74 N

Since the normal force is equal to the force due to kinetic friction (F_friction) in this case, we can use this information to calculate the coefficient of kinetic friction (μ) between the block and the table:

μ = F_friction / N = 48.4 N / 12.74 N = 3.8

Therefore, the coefficient of kinetic friction between the block and the table is 3.8.