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 need to use the principles of projectile motion and equations of motion.
(a) To determine how far up the plane the block goes, we can use the equation for the vertical displacement of an object in projectile motion. The vertical displacement (Δy) can be calculated using the following equation:
Δy = v0 * sin(θ) * t - (1/2) * g * t^2
Where:
v0 = initial velocity of the block = 8.42 m/s
θ = angle of incline = 58.4°
t = time taken to reach the highest point
g = acceleration due to gravity = 9.8 m/s^2
Since the block reaches its highest point when its vertical velocity becomes zero, we can find the time taken to reach the highest point by using the equation:
v = v0 * sin(θ) - g * t
Setting v = 0, we solve the equation for t:
0 = v0 * sin(θ) - g * t
t = v0 * sin(θ) / g
Substituting this value of t into the equation for Δy, we get:
Δy = (v0 * sin(θ) * (v0 * sin(θ) / g)) - (1/2) * g * (v0 * sin(θ) / g)^2
Simplifying the equation gives us the vertical displacement.
(b) To determine the time it takes to reach the highest point, we already calculated it as t = v0 * sin(θ) / g in part (a).
(c) To find the speed of the block when it gets back to the bottom of the plane, we can use the equation for the horizontal velocity of an object in projectile motion. The horizontal velocity (vx) remains constant throughout the motion. Thus, we can use the equation:
vx = v0 * cos(θ)
Therefore, the speed (v) when the block reaches the bottom is equal to the magnitude of vx:
v = |vx| = |v0 * cos(θ)|
Now, let's plug in the given values and calculate each part of the question:
v0 = 8.42 m/s
θ = 58.4°
g = 9.8 m/s^2
(a) Using the equation Δy = v0 * sin(θ) * t - (1/2) * g * t^2, we can calculate Δy.
(b) Using the equation t = v0 * sin(θ) / g, we can calculate t.
(c) Using the equation v = |v0 * cos(θ)|, we can calculate v.
To solve this problem, we will use the equations of motion for an object on an inclined plane.
(a) To find how far up the plane the block goes, we need to find the vertical displacement. We can use the formula:
Δy = (v0^2 * sin^2(θ)) / (2 * g)
Where:
Δy = Vertical displacement
v0 = Initial speed = 8.42 m/s
θ = Angle of incline = 58.4°
g = Acceleration due to gravity = 9.8 m/s²
Substituting the given values into the equation:
Δy = (8.42^2 * sin^2(58.4°)) / (2 * 9.8)
Calculating this expression gives:
Δy ≈ 4.415 m
Therefore, the block goes approximately 4.415 meters up the plane.
(b) To find the time it takes to reach that height, we can use the equation:
Δy = (v0 * sin(θ) * t) - (1/2 * g * t^2)
Rearranging the equation to solve for time (t):
Δy + (1/2 * g * t^2) = v0 * sin(θ) * t
t^2 * (1/2 * g) - (v0 * sin(θ) * t) + Δy = 0
This equation is a quadratic equation with respect to time, t. We can solve it using the quadratic formula:
t = (-b ± sqrt(b^2 - 4 * a * c)) / (2 * a)
Where:
a = 1/2 * g
b = -v0 * sin(θ)
c = Δy
Substituting the given values into the equation:
a = 1/2 * 9.8 = 4.9
b = -8.42 * sin(58.4°)
c = 4.415
Using the quadratic formula:
t = (-(-8.42 * sin(58.4°)) ± sqrt((-8.42 * sin(58.4°))^2 - 4 * (1/2 * 9.8) * 4.415)) / (2 * (1/2 * 9.8))
Calculating this expression gives two possible values for t:
t ≈ 1.151 s (ignoring the negative solution)
Therefore, it takes approximately 1.151 seconds for the block to reach that height.
(c) To find the speed when it gets back to the bottom, we can use the equation:
vf = sqrt(v0^2 + 2 * a * d)
Where:
vf = Final velocity
v0 = Initial speed = 8.42 m/s
a = Acceleration due to gravity = 9.8 m/s²
d = Vertical displacement = 4.415 m (from part a)
Substituting the given values into the equation:
vf = sqrt(8.42^2 + 2 * 9.8 * 4.415)
Calculating this expression gives:
vf ≈ 13.047 m/s
Therefore, the speed of the block when it gets back to the bottom is approximately 13.047 m/s.