Problem 2- A 350 kg roller coaster starts from rest at point A and slides down the frictionless loop-the-loop shown in the figure.

a) How fast is the roller coaster moving at point B?
b) How hard does it press against the track at point B?

A=mgh

= 350 x 9.8 x 25
= 85750 J
B=MGH
= 350 x 9.8 x 12
= 41160 J

1/2mv^2= A-B
v=√(2(A-B)/m)
=16 m/s

F= (mv^2)/r - mg
= 14933.3333 - 3430
= 11503.333 N or simply 1.15 x 10^4 N

Use energy principles on A.

b. force=mg+mv^2/r

a) How fast is the roller coaster moving at point B?

Well, the roller coaster has been on quite a ride, so let's make sure it doesn't break any speed limits! To find the speed at point B, we can consider the conservation of mechanical energy. Since there is no friction, the only forces at play are gravity and the normal force.

Using the law of conservation of energy, we know that the initial potential energy is equal to the final kinetic energy. At point A, the roller coaster has only potential energy, given by PE = mgh, where m is the mass, g is the acceleration due to gravity, and h is the height.

At point B, the roller coaster has both potential energy and kinetic energy. The potential energy is now zero because it's at the bottom of the loop, so we have KE = 1/2mv^2, where v is the speed at point B.

Therefore, equating the potential energy at point A with the kinetic energy at point B, we have:

mgh = 1/2mv^2

Canceling out the mass, m, and solving for v, we get:

v = sqrt(2gh)

So, to find the speed at point B, you just need to take the square root of 2 times the acceleration due to gravity times the height. Just be careful not to get dizzy!

b) How hard does the roller coaster press against the track at point B?

Ah, the roller coaster and its track have quite the complicated "relationship." At point B, the track has to exert a force on the roller coaster to keep it from falling off. This force is called the normal force.

Luckily for us, at the bottom of the loop, the normal force can be found rather easily. The net force experienced by the roller coaster at point B is the centripetal force, which is provided by the normal force.

The centripetal force is given by Fc = mv^2/r, where m is the mass of the roller coaster, v is its velocity, and r is the radius of the loop-the-loop.

So, the normal force is the force exerted by the track in the upward direction, opposite to the downward force of gravity. This normal force can be found by subtracting the force of gravity from the centripetal force:

Fn = Fc - mg

Remember, the normal force has to balance out the forces acting on the roller coaster to keep it on the track. So, if you calculate the centripetal force and subtract the force of gravity, you'll find the answer to how hard the roller coaster presses against the track at point B. But don't worry, the roller coaster is built tough and can handle it!

To determine the speed of the roller coaster at point B and the force exerted on it by the track, we'll need to use principles of conservation of energy and centripetal force.

a) To find the speed of the roller coaster at point B, we'll use the principle of conservation of energy.

At point A, we can assume that all the potential energy is converted into kinetic energy at point B.
The potential energy at point A is given by:
PE_A = m * g * h_A

Where:
m = mass of the roller coaster = 350 kg
g = acceleration due to gravity = 9.8 m/s^2
h_A = height at point A

Since the roller coaster is at rest at point A, its initial kinetic energy is zero.

Therefore, at point B, the sum of the potential energy and kinetic energy should be equal to the potential energy at point A:

PE_A = KE_B

Therefore:

m * g * h_A = (1/2) * m * v_B^2

Simplifying the equation, we get:

v_B = sqrt(2 * g * h_A)

b) To find the force exerted by the track on the roller coaster at point B, we'll use the concept of centripetal force.

The force exerted by the track at point B provides the necessary centripetal force to keep the roller coaster moving in a circular path.
The centripetal force is given by:

F_c = (m * v_B^2) / r

Where:
m = mass of the roller coaster = 350 kg
v_B = velocity at point B (calculated in part a) = sqrt(2 * g * h_A)
r = radius of the circular path at point B

Note: In this case, h_A represents the height of the loop-the-loop at point A, and it determines the radius of the circular path at point B.

By using the conservation of energy principle and calculating the centripetal force, we can find the speed of the roller coaster at point B and the force exerted by the track at point B.

To solve this problem, we can apply the principles of conservation of energy and centripetal force.

a) To find the speed of the roller coaster at point B, we'll use the principle of conservation of energy. At point A, the roller coaster has gravitational potential energy (due to its height) and no kinetic energy (since it's at rest). At point B, all of the potential energy has been converted to kinetic energy.

Using the conservation of energy equation:

mgh = 1/2 mv^2

where m is the mass of the roller coaster (350 kg), g is the acceleration due to gravity (9.8 m/s^2), h is the height from point A to B, and v is the speed at point B.

From the figure, it appears that the height h is slightly higher than the radius of the loop-the-loop. We'll assume the height is 10 meters for this example.

Substituting the known values into the equation:

(350 kg)(9.8 m/s^2)(10 m) = 1/2 (350 kg) v^2

Simplifying the equation:

3430 = 1/2 v^2

Multiplying both sides by 2:

6860 = v^2

Taking the square root of both sides:

v ≈ 82.7 m/s

Therefore, the roller coaster is moving at approximately 82.7 m/s at point B.

b) To find the force that the roller coaster exerts on the track at point B, we'll use the concept of centripetal force. At the top of the loop, the force of gravity provides the centripetal force. Therefore, the force exerted by the roller coaster on the track at point B is equal to its weight.

The weight of the roller coaster is given by:

Weight = mass * acceleration due to gravity

Weight = (350 kg) * (9.8 m/s^2)

Weight ≈ 3430 N

Therefore, the roller coaster exerts a force of approximately 3430 Newtons on the track at point B.