The roller-coaster car shown in Fig. 6-41 (h1 = 50 m, h2 = 11 m, h3 = 30 m), is dragged up to point 1 where it is released from rest. Assuming no friction, calculate the speed at points 2, 3, and 4

you are mean

do 72=9D

8=D

use the formula .5xd*v^2. Use this and you should get all three correct

To calculate the speed at points 2, 3, and 4 on the roller-coaster, we can make use of the principle of conservation of energy. According to this principle, the total mechanical energy of the system is conserved, assuming no non-conservative forces (like friction) are acting.

The total mechanical energy of the system can be divided into two components: kinetic energy (KE) and potential energy (PE). The kinetic energy is due to the motion of the roller coaster car, while the potential energy is due to its vertical position.

At point 1, the roller coaster car is released from rest, meaning it has no initial kinetic energy (KE1 = 0). However, it does have potential energy due to its height from the ground, given by PE1 = mgh1, where m is the mass of the car, g is the acceleration due to gravity, and h1 is the height.

At point 2, the car has descended from the initial height h1 to a lower height h2. At this point, it will have some potential energy and some kinetic energy. The total mechanical energy at point 2 is the sum of the potential and kinetic energies, which is given by:

E2 = KE2 + PE2

Since the roller coaster car is moving, it will have some kinetic energy (KE2 = 1/2mv2), where v2 is the velocity at point 2. The potential energy at this point is given by PE2 = mgh2.

Similarly, at points 3 and 4, we can calculate the total mechanical energy using the same approach. At point 3, the potential energy is given by PE3 = mgh3, and at point 4, the potential energy is given by PE4 = mgh4 (where h4 is the final height at point 4).

However, note that at point 4, the roller coaster car is at its lowest position, meaning it has the maximum kinetic energy and zero potential energy. Thus, we have KE4 = KEmax = 1/2mv4^2, where v4 is the velocity at point 4.

To find the velocities at points 2, 3, and 4, we can equate the sum of the kinetic and potential energies at each point:

E1 = E2
mgh1 = 1/2mv2^2 + mgh2

E2 = E3
1/2mv2^2 + mgh2 = 1/2mv3^2 + mgh3

E3 = E4
1/2mv3^2 + mgh3 = 1/2mv4^2

By rearranging these equations and solving for v2, v3, and v4, we can find the velocities at points 2, 3, and 4, respectively.

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