A block is given an initial velocity of 5.00 m/s up a frictionless 20degree incline. How far up the incline does the block slide before coming to rest?

Hey there, I did the same question but I got 3.78m as my final answer. I used the same calculation and values that you did too.

Yes my answer is also 3.73

The solution is correct but he/she used radians instead of degrees

To find the distance the block slides up the incline before coming to rest, we can use the equations of motion.

The initial velocity of the block is given as 5.00 m/s. We need to find the distance the block travels before coming to rest, so we should focus on the final velocity.

In this case, the final velocity will be zero because the block comes to rest. Since there is no acceleration in the horizontal direction (frictionless condition), we can use the formula for the final velocity in horizontal motion:

V_fx = V_ix + a_xt

where V_fx is the final velocity in the x-direction, V_ix is the initial velocity in the x-direction, a_x is the acceleration in the x-direction, and t is the time taken.

In this problem, the block is moving up the incline, so we need to decompose the initial velocity into its x-direction and y-direction components.

The x-direction component of the initial velocity (V_ix) can be found using trigonometry:

V_ix = V_i * cos(θ)
= 5.00 m/s * cos(20°)

Similarly, the y-direction component of the initial velocity (V_iy) can be found:

V_iy = V_i * sin(θ)
= 5.00 m/s * sin(20°)

Since there is no acceleration in the horizontal direction (a_x = 0 m/s²) and the final velocity is zero (V_fx = 0 m/s), we can rearrange the equation to solve for time:

t = (V_fx - V_ix) / a_x

Plugging in the known values:

t = (0 m/s - V_ix) / 0 m/s²

Now we have the time taken for the block to come to rest in the x-direction (horizontal).

Next, we can use the equation of motion to find the distance traveled by the block in the y-direction (vertical):

Δy = V_iy * t + (1/2) * a_y * t²

Since the incline is at an angle of 20° to the horizontal, the acceleration in the y-direction (a_y) can be calculated using:

a_y = g * sin(θ)

where g is the acceleration due to gravity (about 9.8 m/s²).

Plugging in the known values:

a_y = 9.8 m/s² * sin(20°)

Now substitute the values of V_iy, t, and a_y into the above equation to find the distance traveled (Δy).

Δy = V_iy * t + (1/2) * a_y * t²

This will give us the distance up the incline that the block slides before coming to rest.

If you got 3.78, chances are you dropped the 2 in the denominator. 2a, not just a. 1.4 is correct.

here

vf =0
vo = 5m/s
a=g.sin(theta)
theta=20 degree
g=-9.8 m/s^2
s=?
use

s= [vf^2 - vo^2]/2a

Ans=1.4 m