A cart rolls down a 2.5 m frictionless ramp in 1.25s after starting from rest at the top. What is the angle of the ramp?
a = g sinθ
s = 1/2 at^2
so, now you have
1/2 * g sinθ * 1.25^2 = 2.5
sinθ = 2.5/7.66
θ = 19°
Well, it sounds like the cart really knows how to roll with it! To figure out the angle of the ramp, we can use some good ol' physics.
First, let's break it down. The cart rolled down a distance of 2.5 m in a time of 1.25 s. Now, we can use the kinematic equation:
d = (1/2)at^2
where "d" is the distance, "a" is the acceleration, and "t" is the time. Since the cart was starting from rest, the initial velocity (v₀) is 0.
So, the equation becomes:
2.5 = (1/2) * a * (1.25)^2
Now, let's solve for the acceleration:
a = (2 * 2.5) / (1.25)^2
a = 10 / 1.5625
a = 6.4 m/s²
Since the cart is rolling down a frictionless ramp, the only force acting on it is its weight. The weight force (W) can be represented as:
W = m * g * sin(θ)
where "m" is the mass of the cart, "g" is the acceleration due to gravity (9.8 m/s²), and "θ" is the angle of the ramp.
Since the cart's mass isn't given, we can cancel it out by dividing both sides of the equation by "m":
g * sin(θ) = a
Substituting the known values, we have:
9.8 * sin(θ) = 6.4
Now, let's solve for the angle of the ramp:
sin(θ) = 6.4 / 9.8
θ ≈ 40.1°
Therefore, the angle of the ramp is approximately 40.1 degrees. So, it seems like the cart took a bit of a downhill ride, but it made it look easy-peasy!
To find the angle of the ramp, we can use the formula for the acceleration due to gravity along the ramp.
The formula for the acceleration down a ramp without friction is given by:
a = g * sin(theta)
Where:
a is the acceleration along the ramp,
g is the acceleration due to gravity (approximately 9.8 m/s^2), and
theta is the angle of the ramp.
Since the cart starts from rest, the initial velocity (u) is 0 m/s.
To find the acceleration (a), we can use the second equation of motion:
a = (v - u) / t
Where:
v is the final velocity (which we will assume is the velocity at the bottom of the ramp),
u is the initial velocity (which is 0 m/s),
and t is the time taken to reach the bottom of the ramp (1.25 s).
Given that the cart rolls a distance (s) of 2.5 m, we can use the first equation of motion:
s = ut + (1/2) * a * t^2
Substituting the values into the equation:
2.5 = 0 + (1/2) * a * (1.25)^2
Simplifying the equation:
2.5 = (1.5625/2) * a
Multiplying both sides by 2:
5 = 1.5625 * a
Dividing both sides by 1.5625:
a = 5 / 1.5625 = 3.2 m/s^2
Now, let's substitute this value of acceleration into the first equation to find the angle (theta):
a = g * sin(theta)
3.2 = 9.8 * sin(theta)
Dividing both sides by 9.8:
sin(theta) ≈ 3.2 / 9.8 = 0.3265
To find the angle (theta), we can take the inverse sine (also known as arcsin) of 0.3265:
theta ≈ arcsin(0.3265)
Using a calculator, the angle is approximately 19.14 degrees.
Therefore, the angle of the ramp is approximately 19.14 degrees.
To find the angle of the ramp, we need to use the equation that relates the distance traveled along the ramp, the acceleration, and the angle of the ramp.
First, we need to find the acceleration of the cart. We can use the equation:
\[ d = \frac{1}{2} a t^2 \]
Where:
- \(d\) is the distance traveled along the ramp (2.5 m)
- \(a\) is the acceleration
- \(t\) is the time taken to travel the distance (1.25 s)
Rearranging the equation to solve for acceleration:
\[ a = \frac{2 \cdot d}{t^2} \]
Substituting the given values into the equation:
\[ a = \frac{2 \cdot 2.5\,m}{(1.25\,s)^2} \]
\[ a = \frac{2 \cdot 2.5}{1.25^2}\,m/s^2 \]
\[ a = \frac{5}{1.5625}\,m/s^2 \]
\[ a \approx 3.20\,m/s^2 \]
Now that we have the acceleration, we can use the following equation to find the angle of the ramp:
\[ a = g \cdot \sin(\theta) \]
Where:
- \( g \) is the acceleration due to gravity (9.8 m/s^2)
- \( \theta \) is the angle of the ramp
Rearranging the equation to solve for the angle:
\[ \theta = \arcsin\left(\frac{a}{g}\right) \]
Substituting the values into the equation:
\[ \theta = \arcsin\left(\frac{3.20\,m/s^2}{9.8\,m/s^2}\right) \]
\[ \theta = \arcsin(0.326) \]
\[ \theta \approx 18.8^\circ \]
Therefore, the angle of the ramp is approximately 18.8 degrees.