Answer:
The minimum speed is 14.53 m/s.
Explanation:
Given that,
r = 11 m
Friction coefficient = 0.51
Suppose we need to find the minimum speed, that the cylinder must make a person move at to ensure they will stick to the wall
When frictional force becomes equal to or greater than the weight of person
Then, he sticks to the wall
We need to calculate the minimum speed
Using formula for speed

Where,


Put the value into the formula


Hence, The minimum speed is 14.53 m/s.
(1) You must find the point of equilibrium between the two forces,
<span>G * <span><span><span>MT</span><span>ms / </span></span><span>(R−x)^2 </span></span>= G * <span><span><span>ML</span><span>ms / </span></span><span>x^2
MT / (R-x)^2 = ML / x^2
So,
x = R * sqrt(ML * MT) - ML / (MT - ML)
R = is the distance between Earth and Moon.
</span></span></span>The result should be,
x = 3.83 * 10^7m
from the center of the Moon, and
R - x = 3.46*10^8 m
from the center of the Earth.
(2) As the distance from the center of the Earth is the number we found before,
d = R - x = 3.46*10^8m
The acceleration at this point is
g = G * MT / d^2
g = 3.33*10^-3 m/s^2
As stated in the statement, we will apply energy conservation to solve this problem.
From this concept we know that the kinetic energy gained is equivalent to the potential energy lost and vice versa. Mathematically said equilibrium can be expressed as


Where,
m = mass
= initial and final velocity
g = Gravity
h = height
As the mass is tHe same and the final height is zero we have that the expression is now:






Answer:
A force is a push or pull upon an object resulting from the object's interaction with another object. Whenever there is an interaction between two objects, there is a force upon each of the objects. ... Forces only exist as a result of an interaction.