A small object of mass m = 2.25 kg is released, starting at rest, from a height h above the ground on a ramp inclined at 45.0o (see figure below). The object reaches the bottom of the ramp and

enters a loop-the-loop of radius R = 1.25 m. There is no friction between the object and the track. Treat the object as a point mass.

a) What is the speed of the object at position B, the top of the loop, if h = 5.00 m?
b) What is the magnitude of the normal force acting on the object when it is at position B,
the top of the loop, having started at the height h = 5.00 m? Be sure to draw a free body
diagram for m at the top of the loop.
c) Find the minimum starting height, hmin, for which the object will just make it through the
loop without leaving the track at B. (Hint: Consider what happens to the normal force at the top of the loop when the object just makes it through the loop).

a) To find the speed of the object at position B, we can make use of the conservation of energy principle.

The initial potential energy of the object at height h is given by mgh, where m is the mass, g is the acceleration due to gravity, and h is the height.

At the bottom of the ramp, the object has converted all its potential energy to kinetic energy, which is given by (1/2)mv^2, where v is the velocity.

In the absence of friction, mechanical energy is conserved. Therefore, the mechanical energy at position B is equal to the mechanical energy at the start.

We can equate the initial potential energy to the kinetic energy at position B:

mgh = (1/2)mv^2

Since mass can be canceled from both sides of the equation, we can solve for v:

gh = (1/2)v^2

v^2 = 2gh

Taking the square root of both sides, we get:

v = √(2gh)

Plugging in the given values, with h = 5.00 m and g ≈ 9.8 m/s^2, we can calculate the speed at position B.

b) To determine the magnitude of the normal force at position B, we need to consider the forces acting on the object at the top of the loop.

At this point, the object is moving in a circular path, and there are two forces acting on it: the gravitational force (mg) and the normal force (N).

The gravitational force is directed downward, and the normal force is directed towards the center of the circular path.

At the top of the loop, the normal force is the only force providing the centripetal force required to keep the object moving in a circular path.

By applying Newton's second law in the vertical direction, we can equate the net force to the mass times the acceleration:

N - mg = mv^2 / R

Where R is the radius of the loop.

Solving for N, we get:

N = mg + (mv^2 / R)

Plugging in the known values, such as mass (m), gravitational acceleration (g), velocity (v), and radius (R), we can calculate the magnitude of the normal force at position B.

c) The minimum starting height, hmin, is the height at which the object just makes it through the loop without leaving the track at position B.

To determine this, we need to consider the normal force at the top of the loop.

When the object just makes it through the loop, the normal force at the top is zero, as it is the minimum value required for the object to remain in contact with the track.

Using the equation from part b, when the normal force (N) is zero:

mg + (mv^2 / R) = 0

Solving for the minimum velocity required to just make it through the loop, we get:

vmin^2 = -Rg

To find the minimum starting height, we need to find the potential energy at that height.

mg(hmin - R) = (1/2)mvmin^2

Substituting vmin^2 from the previous equation, we get:

mg(hmin - R) = -Rmg/2

Simplifying, we find:

hmin = R/2

Plugging in the given radius (R = 1.25 m), we can calculate the minimum starting height (hmin) for which the object will just make it through the loop without leaving the track.

a. V^2 = Vo^2 + 2g*h.

V^2 = 0 + 19.6*5 = 98,
V = 9.90 m/s.

b. Fn = Mg*Cos A = 2.25*9.8*Cos 0 = 22.1 N. = Normal force.