A block slides down a frictionless plane having an inclination of 14.3°.

If the block starts from rest at the top and the length of this incline is 1.77 m, find the acceleration of the block.

b)What is the block's speed when it reaches the bottom of the incline?

i found the acceleration, but i don't know how to find the final velocity. is there a formula that i should use?

If the acceleration is "a", the speed V after travelling a distance X, after starting from rest, is

V = sqrt(2aX)

You can derive this yourself using
V = a t
X = (1/2) a t^2
t = sqrt (2X/a)
and substituting for t in the V = at equation.

To find the acceleration of the block sliding down the inclined plane, you can use the equation:

a = g * sin(theta)

where "a" is the acceleration, "g" is the acceleration due to gravity (approximately 9.8 m/s^2), and "theta" is the angle of the incline (14.3° in this case).

Substituting the values into the equation, we get:

a = 9.8 m/s^2 * sin(14.3°)

Calculating this, we find that the acceleration is approximately 2.53 m/s^2.

To find the final velocity of the block when it reaches the bottom of the incline, you can apply the kinematic equation:

v^2 = u^2 + 2as

where "v" is the final velocity, "u" is the initial velocity (which is 0 m/s in this case since the block starts from rest), "a" is the acceleration, and "s" is the distance traveled.

In this case, the distance traveled is the length of the incline, which is given as 1.77 m.

Substituting the values into the equation, we get:

v^2 = 0^2 + 2 * 2.53 m/s^2 * 1.77 m

Simplifying this, we find:

v^2 = 8.9646 m^2/s^2

Taking the square root of both sides, we obtain:

v = √(8.9646) m/s

Calculating this, we find that the speed of the block when it reaches the bottom of the incline is approximately 2.99 m/s.