A block is pushed up a frictionless inclinded plane with initial speed V0 =5.50 m/s. The angle of incline is 0= 23.0º. How far up the plane does the block go? How long does it take to get there?
KE=PE
mv₀²/2=mgh=mgs/sinα
s= v₀²sinα /2g=
=5.5²sin23°/2•9.8 =…
v=v₀-at
v=0 => a= v₀/t
s=v₀t-at²/2=
= v₀t- v₀t²/2t= v₀t/2
t=2s/v₀=….
To determine how far up the plane the block goes and how long it takes to get there, we can use the principles of motion equations.
Let's break down the given information:
Initial speed, V0 = 5.50 m/s
Angle of incline, θ = 23.0º
First, let's analyze the motion along the inclined plane using the horizontal (x) and vertical (y) directions.
In the x-direction, the block experiences no acceleration because the plane is frictionless. Therefore, the initial velocity in the x-direction (V0x) remains constant throughout the motion. We can find V0x using the initial speed and the angle of incline:
V0x = V0 * cos(θ)
V0x = 5.50 m/s * cos(23.0º)
V0x ≈ 5.50 m/s * 0.921
V0x ≈ 5.06 m/s
In the y-direction, the block experiences constant acceleration due to gravity (g = 9.8 m/s^2). We need to determine how far up the plane the block goes, which we'll call "h," and the time it takes to reach that point, which we'll call "t."
The displacement along the y-direction (h) can be calculated using the equation:
h = (V0y^2) / (2g)
Since the initial vertical velocity (V0y) is zero (the block starts from rest in the vertical direction), we can simplify the equation to:
h = (V0sin(θ)^2) / (2g)
h = (V0 * sin(θ))^2 / (2 * g)
h = (5.50 m/s * sin(23.0º))^2 / (2 * 9.8 m/s^2)
h ≈ (5.50 m/s * 0.390)^2 / (19.6 m/s^2)
h ≈ 12 m
So, the block goes approximately 12 meters up the plane.
Next, let's find the time it takes for the block to reach that height.
We can use the equation of motion along the y-direction:
h = V0y * t + (1/2) * g * t^2
Since the displacement (h) is known and the initial vertical velocity (V0y) is zero, we're left with:
h = (1/2) * g * t^2
Solving for t:
t = √((2h) / g)
t = √((2 * 12 m) / 9.8 m/s^2)
t ≈ √(2.448 m / 9.8 m/s^2)
t ≈ √(0.249 s^2)
t ≈ 0.499 s
Therefore, it takes approximately 0.499 seconds for the block to reach the height of 12 meters up the incline.