Consider a roller coaster that moves along the track shown in the figure below. Assume all friction is negligible and ignore the kinetic energy of the roller coaster's wheels. Assume the roller coaster starts with a speed of

19 m/s at point A. Find the speed of the roller coaster when it reaches locations B and C.(Assume hA = 40 m and hB = 20 m.)
VB=____m/s
VC=____m/s

To find the speed of the roller coaster at points B and C, we can use the conservation of mechanical energy principle. According to this principle, the total mechanical energy of the roller coaster remains constant throughout the motion if we neglect friction.

The total mechanical energy (E) of the roller coaster is given by the sum of its potential energy (PE) and kinetic energy (KE):

E = PE + KE

At point A, the roller coaster has a height (hA) and a speed (vA):

hA = 40 m (given)
vA = 19 m/s (given)

The total mechanical energy at point A can be calculated as:

EA = PE + KE = mghA + 0.5mvA^2

Now, let's find the total mechanical energy at points B and C using the fact that the roller coaster has heights hB = 20 m and hC = 0 m, respectively.

At point B:
EB = mghB + 0.5mvB^2

At point C:
EC = mghC + 0.5mvC^2

Since the total mechanical energy is conserved, we have:

EA = EB = EC

Therefore, we can equate the expressions for the total mechanical energy at each point:

mghA + 0.5mvA^2 = mghB + 0.5mvB^2 = mghC + 0.5mvC^2

Simplifying this equation, we can find vB and vC:

hA + 0.5vA^2 = hB + 0.5vB^2 = hC + 0.5vC^2

Substituting the given values, we get:

40 + 0.5(19^2) = 20 + 0.5vB^2 = 0 + 0.5vC^2

Solving the equation, we find:

400 + 0.5vB^2 = 0.5vC^2

vB^2 = vC^2 - 800

Since we don't have a specific value for vB or vC, we can't directly solve for them. However, we can determine the relationship between vB and vC.

From the equation above, we see that vB^2 will be greater than vC^2 by 800. This means that vB is greater than vC. Therefore, we can conclude:

VB > VC

In summary, we cannot determine the exact values of vB and vC without specific information about the track and the roller coaster's mass. However, we know that the speed at point B (VB) will be greater than the speed at point C (VC).

To find the speed of the roller coaster at locations B and C, we can use the principle of conservation of mechanical energy. This principle states that the sum of the roller coaster's potential energy and kinetic energy at any given point along the track remains constant, assuming no energy losses due to friction.

Let's start by calculating the potential energy of the roller coaster at points A, B, and C, respectively.

The potential energy (PE) of an object is given by the equation:

PE = mgh

Where m is the mass of the roller coaster, g is the acceleration due to gravity (approximately 9.8 m/s^2), and h is the height above a reference point.

Given that hA = 40 m and hB = 20 m, we can calculate the potential energy at points A and B as follows:

PEA = m * g * hA
PEB = m * g * hB

Now, at point A, the roller coaster has both potential energy and kinetic energy. The total mechanical energy (E) at point A is given by:

E = PEA + KEA

where KEA is the kinetic energy of the roller coaster at point A.

Since the problem states that the roller coaster starts with a speed of 19 m/s at point A, we can calculate the kinetic energy using the formula:

KEA = 0.5 * m * vA^2

where vA is the initial velocity at point A (19 m/s).

Now, let's find the velocity at point B. At this point, the roller coaster only has kinetic energy, as it has descended the hill and lost all its potential energy.

Using the principle of conservation of energy, we can equate the total mechanical energy at point A to the kinetic energy at point B:

E = KEA = 0.5 * m * vB^2

Solving for vB gives us the speed of the roller coaster at point B.

Similarly, we can repeat the same process for point C. At point C, the roller coaster loses all its kinetic energy, so its total mechanical energy is equal to the potential energy at point C:

E = PEC = m * g * hC

Solving for vC gives us the speed of the roller coaster at point C.

Therefore,

VB = sqrt(2 * (E - m * g * hB) / m)
VC = sqrt(2 * (E - m * g * hC) / m)

where E = PEA + KEA.

By substituting the given values of hA, hB, and vA, and solving for VB and VC using the equations above, we can find the speeds of the roller coaster at points B and C.

KE(A)+PE(A)=KE(B)

mv(A)²/2 +mgh(A)= mv(B)²/2
v(B)=sqrt[v(A)²+2gh(A)] =
=sqrt[19²+2•9.8•40) =33.84 m/s
KE(B)+PE(B)=KE(C)
mv(B)²/2 +mgh(B)= mv(C)²/2
v(C)=sqrt[v(B)²+2gh(B)] =
=sqrt[33.84²+2•9.8•20) =43.92 m/s