A block is given an initial velocity of 9.00 m/s up a frictionless incline of angle θ = 21.0°. How far up the incline does the block slide before coming to rest?

To find out how far up the incline the block slides before coming to rest, we need to consider the forces acting on the block and the work done by those forces.

In this case, the only force acting on the block is its weight (mg), which can be split into two components: one parallel to the incline (mg*sinθ) and one perpendicular to the incline (mg*cosθ).

The force parallel to the incline (mg*sinθ) is responsible for the block's acceleration and will eventually cause it to come to rest. This force can be equated to the negative of the force of kinetic friction (μk * N), where μk is the coefficient of kinetic friction and N is the normal force. Since the incline is frictionless, the force of kinetic friction is zero.

Therefore, we have:

mg*sinθ = 0

Now, we can solve for the distance traveled up the incline.

Using the equation of motion, v^2 = u^2 + 2as, where v is the final velocity (which is zero in this case), u is the initial velocity (9.00 m/s), a is the acceleration along the incline (g*sinθ), and s is the distance traveled.

Plugging in the values, we have:

0^2 = (9.00 m/s)^2 + 2 * (g*sinθ) * s

Rearranging the equation, we get:

s = - (9.00 m/s)^2 / (2 * g * sinθ)

Now, we can substitute the values of g (acceleration due to gravity) and θ (angle of the incline) to find the distance traveled.

Using g ≈ 9.81 m/s^2 and θ = 21.0°, we get:

s = - (9.00 m/s)^2 / (2 * 9.81 m/s^2 * sin(21.0°))

Calculating this expression gives us:

s ≈ -5.64 m

Since distance cannot be negative in this context, the block slides approximately 5.64 meters up the incline before coming to rest.