A 2.90 kg block starts from rest at the top of a 30° incline and accelerates uniformly down the incline, moving 2.23 m in 1.90 s.

a) Find the magnitude of the acceleration of the block.

(b) Find the coefficient of kinetic friction between the block and the incline.

(c) Find the magnitude of the frictional force acting on the block.

(d) Find the speed of the block after it has slid a distance 2.23 m.

To solve this problem, we can use the equations of motion along with the principles of Newtonian mechanics. Let's break down each part of the problem and explain the steps to solve it.

(a) Find the magnitude of the acceleration of the block:

We can start by using the equation of motion: s = ut + (1/2)at^2, where s is the displacement, u is the initial velocity, a is the acceleration, and t is the time.

Given information:
- The displacement (s) = 2.23 m
- The initial velocity (u) = 0 m/s (as the block starts from rest)
- The time (t) = 1.90 s

Rearranging the equation, we can solve for acceleration (a):

s = ut + (1/2)at^2
2.23 = 0 + (1/2)a(1.90)^2
2.23 = 0.9a
a = 2.23 / 0.9
a ≈ 2.47 m/s^2

Therefore, the magnitude of the acceleration of the block is approximately 2.47 m/s^2.

(b) Find the coefficient of kinetic friction between the block and the incline:

To find the coefficient of kinetic friction, we need to use the following equation of motion: a = g.sin(θ) - μk.g.cos(θ), where g is the acceleration due to gravity and θ is the angle of the incline.

Given information:
- The angle of incline (θ) = 30°
- The acceleration due to gravity (g) = 9.8 m/s^2 (standard value on Earth)

Rearranging the equation, we can solve for the coefficient of kinetic friction (μk):

a = g.sin(θ) - μk.g.cos(θ)
μk.g.cos(θ) = g.sin(θ) - a
μk = (g.sin(θ) - a) / (g.cos(θ))

Plugging in the given values:

μk = (9.8.sin(30°) - 2.47) / (9.8.cos(30°))
μk ≈ 0.144

Therefore, the coefficient of kinetic friction between the block and the incline is approximately 0.144.

(c) Find the magnitude of the frictional force acting on the block:

The magnitude of the frictional force can be determined using the equation: f = μk.N, where f is the frictional force and N is the normal force.

Given that the normal force (N) is equal to the weight (mg), where m is the mass of the block and g is the acceleration due to gravity.

Given information:
- The mass of the block (m) = 2.90 kg
- The acceleration due to gravity (g) = 9.8 m/s^2

Substituting these values into the equation:

f = μk.N
f = μk.m.g

f = (0.144)(2.90)(9.8)
f ≈ 4.14 N

Therefore, the magnitude of the frictional force acting on the block is approximately 4.14 N.

(d) Find the speed of the block after it has slid a distance of 2.23 m:

We can use the equation of motion: v^2 = u^2 + 2as, where v is the final velocity, u is the initial velocity, a is the acceleration, and s is the displacement.

Given information:
- The initial velocity (u) = 0 m/s (block starts from rest)
- The displacement (s) = 2.23 m
- The acceleration (a) = 2.47 m/s^2 (calculated in part a)

Rearranging the equation, we can solve for the final velocity (v):

v^2 = u^2 + 2as
v^2 = 0 + 2(2.47)(2.23)
v^2 = 9.2342
v ≈ 3.04 m/s

Therefore, the speed of the block after it has slid a distance of 2.23 m is approximately 3.04 m/s.