A block is projected up a frictionless inclined plane with initial speed v0 = 8.42 m/s. The angle of incline is è = 58.4°. (a) How far up the plane does it go? (b) How long does it take to get there? (c) What is its speed when it gets back to the bottom?
To solve this problem, we will use the equations of motion for motion along an inclined plane.
Given:
Initial speed, v0 = 8.42 m/s
Angle of incline, θ = 58.4°
(a) How far up the plane does it go?
To find the distance traveled up the plane, we can use the equation:
sin(θ) = (v^2 - v0^2) / (2 * g * d)
where sin(θ) is the sine of the angle of incline, v is the final velocity (zero when it reaches the highest point), g is the acceleration due to gravity (approximately 9.8 m/s²), and d is the distance traveled up the plane.
Rearranging the equation, we get:
d = (v^2 - v0^2) / (2 * g * sin(θ))
Since the final velocity at the highest point is zero, we have:
d = (- v0^2) / (2 * g * sin(θ))
Substituting the given values:
v0 = 8.42 m/s
θ = 58.4°
g ≈ 9.8 m/s²
Calculating:
d = (- (8.42 m/s)^2) / (2 * 9.8 m/s² * sin(58.4°))
d ≈ - 4.696 m/s² / (2 * 9.8 m/s² * 0.847)
d ≈ - 0.244 m / 1.632
d ≈ - 0.149 m
The distance traveled up the plane is approximately 0.149 meters. Note that the negative sign indicates it is moving up the incline in the negative direction.
(b) How long does it take to get there?
To find the time it takes to get there, we can use the equation:
v = v0 + a * t
where v is the final velocity (zero when it reaches the highest point), v0 is the initial velocity, a is the acceleration due to gravity along the inclined plane, and t is the time.
Rearranging the equation, we get:
t = (v - v0) / a
In this case, since the final velocity at the highest point is zero, we have:
t = - v0 / a
Substituting the given values:
v0 = 8.42 m/s
a = g * sin(θ) ≈ 9.8 m/s² * sin(58.4°)
Calculating:
t = - (8.42 m/s) / (9.8 m/s² * sin(58.4°))
t = -0.858 s / (9.8 m/s² * 0.847)
t ≈ -0.095 s
The time it takes to reach the highest point is approximately 0.095 seconds. Note that the negative sign indicates it is moving up the incline in the negative direction.
(c) What is its speed when it gets back to the bottom?
To find the speed when it gets back to the bottom, we can use the equation:
v^2 = v0^2 + 2 * a * d
where v is the final velocity at the bottom, v0 is the initial velocity, a is the acceleration due to gravity along the inclined plane, and d is the distance traveled up the plane (which we found to be approximately -0.149 meters).
Rearranging the equation, we get:
v = sqrt(v0^2 + 2 * a * d)
Substituting the given values:
v0 = 8.42 m/s
a = g * sin(θ) ≈ 9.8 m/s² * sin(58.4°)
d ≈ -0.149 m
Calculating:
v = sqrt((8.42 m/s)^2 + 2 * (9.8 m/s² * sin(58.4°)) * (-0.149 m))
v ≈ sqrt(70.96 m²/s² + 2 * (9.8 m/s² * 0.847) * (-0.149 m))
v ≈ sqrt(70.96 m²/s² + (-0.299 m/s² * 9.8 m²/s²) * (-0.149 m))
v ≈ sqrt(70.96 m²/s² + 0.413 m²/s² * 0.149 m)
v ≈ sqrt(70.96 m²/s² + 0.061 m²/s²)
v ≈ sqrt(71.021 m²/s²)
v ≈ 8.43 m/s
The speed when it gets back to the bottom is approximately 8.43 m/s.
To solve this problem, we can apply the principles of projectile motion and use the equations of motion to determine the answers to the given questions.
First, let’s break down the problem and identify the information provided:
Given:
- Initial speed, v0 = 8.42 m/s
- Angle of incline, è = 58.4°
- The inclined plane is frictionless
Now, let's address each question individually:
(a) How far up the plane does it go?
To determine how far up the plane the block goes, we need to find the vertical distance covered. We can decompose the initial velocity (v0) into its horizontal and vertical components.
Vertical component of initial velocity (vy0) = v0 * sin(è)
Horizontal component of initial velocity (vx0) = v0 * cos(è)
Since the inclined plane is frictionless, there is no acceleration in the horizontal direction. However, in the vertical direction, the only acceleration acting on the block is due to gravity, which is directed downwards. We can use the following kinematics equation to determine the vertical distance covered (h):
h = (vy0^2) / (2 * g)
where g is the acceleration due to gravity (approximately 9.8 m/s^2).
Using the given values, we can plug them into the equation and find h:
h = (8.42^2 * sin^2(58.4°)) / (2 * 9.8)
Solving this equation will give you the height the block reaches up the inclined plane.
(b) How long does it take to get there?
To find the time taken to reach the maximum height, we can use the vertical component of the initial velocity, vy0, and the acceleration due to gravity, g.
The equation we can use here is:
t = vy0 / g
By substituting the given values into this equation, we can calculate the time taken to reach the maximum height.
(c) What is its speed when it gets back to the bottom?
When the block reaches the bottom of the inclined plane, its height will be the same as at the start. Therefore, the final vertical velocity (vf) will have the same magnitude but the opposite direction as the initial vertical velocity (vy0).
To find the final speed (vf), we can use the equation:
vf = √(vfx^2 + vfy^2)
where vfx is the final horizontal velocity (which remains constant) and vfy is the final vertical velocity.
The final vertical velocity can be found by reversing the sign of the initial vertical velocity, i.e., making it negative since it is directed downwards.
By plugging the values into the equation, we can calculate the final speed of the block when it gets back to the bottom.
Remember to always double-check your calculations and units to ensure accurate answers.