A package slides down a 135 m long ramp with no friction. If the package starts from rest at the top and is to have a speed no faster than 19m/s at the bottom, what should be the maximum angle of inclination?
(1/2) M Vmax^2 = M*g*135 sinA
M cancels out.
sinA = (1/2)Vmax^2/(g*135)
= 0.1364
A = 7.84 degrees
Why did the package go to therapy? Because it had an inclination for speed!
In order to find the maximum angle of inclination, let's use some physics. We can start by using the equation that relates the speed at the bottom of a ramp to its angle of inclination:
V = √(2 * g * h * sin(θ))
Where:
V = package's speed at the bottom
g = acceleration due to gravity (~9.8 m/s²)
h = height of the ramp
θ = angle of inclination
Given that the package should have a speed no faster than 19 m/s and the height of the ramp is 135 m, we can rearrange the equation to solve for the angle of inclination:
θ = arcsin((V²) / (2 * g * h))
Plugging in the values:
θ = arcsin((19²) / (2 * 9.8 * 135))
Calculating this using a calculator, the maximum angle of inclination should be approximately 25.4 degrees. So, the ramp should have an angle no steeper than that to keep the package's speed as desired.
To find the maximum angle of inclination, we can use the principle of conservation of energy. The initial potential energy at the top of the ramp will be converted into kinetic energy at the bottom of the ramp.
The potential energy (PE) at the top is given by the formula:
PE = mgh
Where m is the mass of the package, g is the acceleration due to gravity (approximately 9.8 m/s²), and h is the vertical height of the ramp.
The kinetic energy (KE) at the bottom is given by the formula:
KE = (1/2)mv²
Where m is the mass of the package and v is the speed at the bottom.
Since we want the package to have a maximum speed of 19 m/s at the bottom, we can set KE to be equal to the maximum kinetic energy:
KE = (1/2)mv² = (1/2)m(19)² = 180.5m
Setting the potential energy equal to the kinetic energy:
mgh = 180.5m
Simplifying and canceling out the mass:
gh = 180.5
Now, we can solve for the height:
h = 180.5/g
Substituting g = 9.8 m/s²:
h = 180.5/9.8 ≈ 18.47
The maximum height of the ramp is approximately 18.47 meters.
Finally, we can use the trigonometric relationship between the height and the length of the ramp to find the maximum angle of inclination:
sinθ = h/length
θ = arcsin(h/length)
θ = arcsin(18.47/135)
θ ≈ 7.85 degrees
Therefore, the maximum angle of inclination should be approximately 7.85 degrees.
To determine the maximum angle of inclination, we need to consider the conservation of energy. As the package slides down the ramp with no friction, the work done by gravity is transformed into the kinetic energy of the package.
Let's break down the problem step by step:
1. Identify the given information:
- Length of the ramp (s): 135 m
- Maximum final velocity (v): 19 m/s
- Acceleration due to gravity (g): 9.8 m/s²
2. Determine the initial potential energy of the package at the top of the ramp:
- Potential energy (PE) = mgh, where m is the mass, g is the acceleration due to gravity, and h is the height.
- Since the package starts at rest, its initial kinetic energy (KE) is zero.
- Therefore, the initial potential energy (PE) is equal to the initial total mechanical energy (Ei).
3. Calculate the final kinetic energy at the bottom of the ramp:
- Since the package has its maximum final velocity at the bottom, its kinetic energy (KE) is ½mv².
4. Set up an energy conservation equation:
- Initial total mechanical energy (Ei) = Final mechanical energy (Ef)
- PE + KE = KE
- mgh + 0 = ½mv²
5. Simplify the energy conservation equation:
- Cancel out the mass (m) on both sides of the equation:
- gh = ½v²
6. Rearrange the equation to solve for the height (h):
- h = ½v² / g
7. Substitute the given values into the equation:
- h = ½ * (19 m/s)² / (9.8 m/s²)
- Calculate h:
- h ≈ 9.4 m
8. Determine the angle of inclination (θ) using trigonometry:
- tan(θ) = h / s
- θ = arctan(h / s)
9. Substitute the values of h and s into the equation:
- θ = arctan(9.4 m / 135 m)
- Calculate θ:
- θ ≈ 3.96°
Therefore, the maximum angle of inclination for the ramp should be approximately 3.96 degrees in order to ensure that the package does not exceed a speed of 19 m/s at the bottom.