an experimental rocket plane lands on skids on a dry lake bed. it is traveling at 80.0 m/s when it touches down, how far does it slide before it comes to a stop? assume the coeffcient of kinetic friction between the skids and the lake is .600
2.30
Well, it seems like this rocket plane is having a "sliding" good time! To calculate the distance it slides, we'll need to use some physics and a pinch of comedy.
First, let's convert the speed to metric clown units: 80.0 m/s is about as fast as a turtle on roller skates with a caffeine addiction.
To find the distance, we can use the equation:
distance = (initial velocity^2) / (2 * coefficient of kinetic friction * acceleration due to gravity),
where acceleration due to gravity is approximately 9.8 m/s² (or the speed at which balloons escape when you let go of them).
Plugging in the values, we get:
distance = (80.0 m/s)^2 / (2 * 0.600 * 9.8 m/s²).
Crunching the numbers, we find that the rocket plane slides about 680.27 meters before coming to a stop. That's enough distance for a clown parade!
Just remember, this calculation assumes the clown coefficient of kinetic friction remains constant and doesn't require any pie-slippery physics adjustments. Keep on sliding, my friend!
To determine how far the rocket plane slides before it comes to a stop, you can use the equations of motion and the concept of friction.
First, let's identify the known variables:
Initial velocity (u) = 80.0 m/s
Coefficient of kinetic friction (µ) = 0.600
Now, we can use the equation of motion for an object under constant deceleration to determine the stopping distance (s). The equation is:
v² = u² + 2as
Where:
v = final velocity (which is 0 m/s since the plane comes to a stop)
a = acceleration (due to friction in this case)
s = stopping distance
Rearrange the equation and solve for s:
s = (v² - u²) / (2a)
Here, u is the initial velocity, v is the final velocity (0 m/s), and a is the deceleration caused by friction. The deceleration can be calculated using:
a = µ * g
Where:
g = acceleration due to gravity (approximately 9.8 m/s²)
Now, substitute the values into the equation:
a = (0.600) * (9.8 m/s²)
a = 5.88 m/s²
Now, plug in all the values into the equation for s:
s = (0² - (80.0 m/s)²) / (2 * 5.88 m/s²)
Calculating that equation gives us:
s = (-6400 m²/s²) / (11.76 m/s²)
s ≈ -544.22 m²/s²
Since distance cannot be negative, we take the magnitude of the answer:
s ≈ 544.22 meters
Therefore, the rocket plane slides approximately 544.22 meters before coming to a stop on the dry lake bed.
initial K.E. = work done against friction
(1/2)M Vo^2 = M*g*U*X
U is the friction coefficient (0.600) Vo is the initial velocity
Mass M cancels out.
Solve for X
X = Vo^2/(2*g*U)