The initial velocity of a 2.56 kg block sliding down a frictionless inclined plane is found to be 1.13 m/s. Then 1.08 s later, it has a velocity of 3.99 m/s.What is the angle of the plane with respect to the horizontal?

Acceleration = a = (3.99-1.13)/1.08

= 2.648 m/s^2

Force along the plane = M*g sinA = M*a

sinA = a/g = 0.270

A = sin^-1(0.270)= 15.7 degrees

To find the angle of the plane with respect to the horizontal, we can use the principles of kinematics and trigonometry.

Let's consider the motion of the block along the inclined plane. We can break the initial velocity into its horizontal and vertical components. The horizontal component will remain constant, while the vertical component will be affected by gravity. Let's call the angle of the plane θ.

We can start by finding the acceleration of the block along the inclined plane using the equation:

a = (v_f - v_i) / t

Where:
a = acceleration along the incline
v_f = final velocity = 3.99 m/s
v_i = initial velocity = 1.13 m/s
t = time interval = 1.08 s

Substituting the values, we have:

a = (3.99 m/s - 1.13 m/s) / 1.08 s
a = 2.86 m/s / 1.08 s
a ≈ 2.648 m/s²

Now, let's find the vertical acceleration (g) using the equation:

g = a_y = a * sin(θ)

Where:
g = acceleration due to gravity = 9.8 m/s² (approximate value)
a_y = vertical acceleration
θ = angle of the plane

Rearranging the equation, we have:

sin(θ) = a_y / a

Substituting the values, we can solve for sin(θ):

sin(θ) ≈ (2.648 m/s²) / (9.8 m/s²)
sin(θ) ≈ 0.27

Now, we can find the angle θ by taking the inverse sine (sin⁻¹) of 0.27:

θ ≈ sin⁻¹(0.27)
θ ≈ 15.6°

Therefore, the angle of the plane with respect to the horizontal is approximately 15.6 degrees.