0.25 Phy1220: relationships among kinematics, newton's laws, vectors  (Page 5/5)

Validation

We can compute the total mechanical energy at this point as

PEg = m*g*h = 10kg * (9.8m/s^2) * 413m = 40474 joules

This result is close enough to the total mechanical energy at the end of Leg B to validate the computations. In this case, we determined the height usingtime, velocity, and acceleration, and validated that height using work/energy concepts.

State at the end of Leg C

At the completion of Leg C:

• The rocket is at the apex at a height of 413 meters.
• The total mechanical energy is 40470 joules.
• The kinetic energy is 0 because for an instant, the rocket isn't moving.
• The mechanical energy consists totally of gravitational potential energy.

Leg D

Leg D of the trip is fairly simple. The rocket falls for a distance of 413 meters under the influence of the internal gravitational force.

No change in mechanical energy

Once again, because the force is an internal force, the total mechanical energy cannot be changed by the work done by the force. However, the mechanicalenergy can be transformed from potential energy to kinetic energy.

At the instant before the rocket strikes the ground, it must still have a total mechanical energy value of 40470 joules.

Kinetic energy: 40470, potential energy: 0

At the instant before the rocket strikes the ground, all of the mechanical energy has been transformed into kinetic energy. We can use that knowledge tocompute the velocity of the rocket right before it strikes the ground.

KE = 0.5*m*v^2, or

v^2 = KE/(0.5*m), or

v = (KE/(0.5*m))^(1/2) = (40470 joules/(0.5*10kg))^(1/2), or

terminal velocity = v = 90 meters/sec

Thus, the terminal velocity of the rocket when it strikes the ground is 90 meters/sec straight down.

Validation

Let's see if we can validate that result using a different approach. Given the height of the apex and theacceleration of gravity, we can computer the transit time as

413m = 0.5*g*t^2, or

t^2 = 413m/(0.5*g), or

t = (413m/(0.5*g))^(1/2) = (413m/(0.5*9.8m/s^2))^(1/2), or

t = 9.18 seconds

Compute the terminal velocity

Knowing the time to make the trip to the ground along with the acceleration, we can compute the terminal velocity as

v = g * t = (9.8m/s^2)*9.18s = 90 m/s

which matches the terminal velocity arrived at on the basis of work and energy.

State at the end of Leg D

Therefore, at the end of Leg D, the rocket crashes into the ground. However, an instant before the crash,

• The total mechanical energy is 40470 joules.
• The gravitational potential energy is 0.
• The kinetic energy is 40470 joules.
• The velocity is 90 m/s straight down toward the center of the earth.

Do the calculations

I encourage you to repeat the calculations that I have presented in this lesson to confirm that you get the same results. Experiment with the scenarios, making changes, and observing the results of your changes. Make certain that you can explain why your changes behave as they do.

Resources

I will publish a module containing consolidated links to resources on my Connexions web page and will update and add to the list as additional modulesin this collection are published.

Miscellaneous

This section contains a variety of miscellaneous information.

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